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Book_i~D> <4_ 

Copyright^ 0 _ 


COPYRIGHT DEPOSIT. 






































































































































































SHEET-METAL PATTERN DRAFTING 


AND 

SHOP PROBLEMS 


By 

JAMES SV'DAUGHERTY 


College of Industries 
Carnegie Institute of Technology 

Pittsburgh, Pa. 



THE MANUAL ARTS PRESS 

Peoria, Illinois 







COPYRIGHT, 1922 
'James S. Daugherty 
12D 22 



r 

'/V 1 • 


Printed in the United States of America. 

©CI.A6748 2 6 


"D^ 

4 


s?-/? w 


PREFACE 


T HIS book has been prepared as a text for use in vocational 
schools, trade schools, technical schools and high schools 
offering courses in sheet-metal pattern drafting and shop work, 
and for home study by apprentices and sheet-metal workers. 
It meets every requirement as a text book, and is also well 
adapted for reference use by draftsmen, shop foremen, and 
metal workers engaged in laying out patterns for general sheet- 
metal work, heating, ventilating, cornice, skylight, and heavy 
plate work. 

The problems are practical and easily adaptable to varying 
courses in sheet-metal work, and have proved of exceptional 
value where pattern drafting and shop work are correlated and 
taught in a thoro, systematic manner in the best vocational 
and technical schools. 

The subject-matter and method of presentation are the out¬ 
come of many years of teaching and practical experience in the 
various branches of the sheet-metal industry. 

The proper sequence so necessary for successful instruction in 
sheet-metal pattern drafting is an important feature of this 


book; also, each problem presented is drawn to scale or dimen¬ 
sions given, and is of ample size for constructing from metal, 
using a minimum amount of material. 

The descriptions are clear and well organized step by step. 
They stimulate the student to think and reason, and simplify 
the instructor’s work. Numerous illustrative problems are 
worked thruout the text, and a large number of examples for 
the student are given in each chapter. The work is so planned 
that the student can take care of himself to a great extent. 

The demonstrations of these practical problems are rendered 
clear by the illustrations which are photographic reproductions 
of work in which the patterns were developed, transferred to 
metal, and constructed by students in the sheet-metal depart¬ 
ment of the Carnegie Institute of Technology. 

JAMES S. DAUGHERTY. 

College of Industries, 

Carnegie Institute of Technology. 

November, 1921. 


3 



CONTENTS 


PART I 

DRAFTING PRINCIPLES 

Chapter I. Drawing Equipment. 9 

Equipment—Paper—Pencils—Preparation of Paper. 

Chapter II. Practical Geometry. 12 

Geometric Problems—The Helix. 

Chapter III. Practical Pattern Drafting. 24 


PART II 

PARALLEL LINE DEVELOPMENTS 
Chapter IV. Pattern Problems. 25 

Problem 1, Pattern for Two-Piece Elbow in Round Pipe—Problem 
2, Conductor Elbow—Problem 3, Rectangular Conductor Shoe— 
Problem 4, Two-Piece Elbow in an Oblong Pipe—Problem 5, Pipe 
and Roof Flange—Problem 6, Pipe Roof Flange for Ridge of Roof 
•—Problem 7, An Octagonal Pipe Fitting Over the Ridge of a Roof 
—Problem 8, T-Joint Between Pipes of Same Diameter—Problem 
9, T-Joint Between Pipes of Different Diameters—Problem 10, 
T-Joint Between Pipes of Same Diameter at an Angle—Problem 
11, Y-Joint—Problem 12, Two Square Pipes of the Same Size In¬ 
tersecting at an Angle—Problem 13, Two Round Pipes of Unequal 
Diameters that Intersect Irregularly—Problem 14, An Octagonal 
Pipe Intersecting a Square Pipe Obliquely—Problem 15, A Rec¬ 
tangular Pipe Intersecting a Round Pipe Obliquely—Problem 16, 
Four-Piece 90° Elbow—Problem 17, Five-Piece 60° Elbow—- 
Problem 18, Three-Piece 90° Elbow—Problem 19, Profiles of El¬ 
bows. 

Chapter V. Practical Cornice and Gutter Problems .. 42 

Problem 20, Roman Moldings—Problem 21, Square Return Miter 
—Problem 22, Butt Miter—Problem 23, An Oblique Return Miter 


-—Problem 24, Butt Miter Against a Surface Oblique in Plan— 
Problem 25, Eave-Trough Miter—Problem 26, Molded Face Gutter 
—Problem 27, Ogee Gutter—Problem 28, Molded Roof Gutter—- 
Problem 29, Octagonal Gutter—Problem 30, Quarter-Circle Gutter 
—Problem 31, Square Face Miter—Problem 32, Octagon Face 
Miter—Problem 33, Miter Between a Gable and Horizontal Mold¬ 
ing—Problem 34, Panel Miter—Problem 35, Roof Finial—Prob¬ 
lem 36, Square Ventilator—Problem 37, Octagonal Roof Finial—- 
Problem 38, Conductor Head—Problem 39, Pediment Molding 
Mitering on a Wash—Problem 40, Gable Molding with Raked Pro¬ 
file—Problem 41, Broken Pediment—Problem 42, Miter Between 
Moldings of Different Profiles. 


PART III 

RADIAL DEVELOPMENTS 

Chapter VI. Regular Tapering Forms. 64 

Problem 43, Pattern for Cone and Frustum—Problem 44, Pattern 
for a Square Pyramid—Problem 45, Pattern for an Irregular Frus¬ 
tum of a Cone—Problem 46, Pattern for an Irregular Frustum of 
a Hexagonal Pyramid—Problem 47, Conical Gutter Outlet—Prob¬ 
lem 48, Tapering Collar for Roof Having a Double Pitch—Problem 
49, Rectangular Pyramid—Problem 50, Octagonal Pyramid—Prob¬ 
lem 51, Oblong Pitched Cover—Problem 52, Flaring Pan—Problem 
53, Pattern for a Funnel—Problem 54, Round Ventilator Head— 
Problem 55, Fruit-Jar Filler—Problem 56, Elliptical Flaring Pan— 
* Problem 57, Grocer’s Scoop—Problem 58, Tapering Square Pipe 
Intersecting a Vertical Square Pipe—Problem 59, A Cone Inter¬ 
sected by a Cylinder Placed Vertically—Problem 60, A Cone In¬ 
tersected by a Vertical Square Pipe—Problem 61, A Cone Inter¬ 
sected by a Cylinder at Right Angles to Its Axis—Problem 62, A 
Cone Intersected by a Square Pipe Placed in a Horizontal Position 
-—Problem 63, The Frustum of a Cone Intersecting a Cylinder 
Obliquely—Problem 64, The Frustum of a Cone Intersecting a 
Cone of Unequal Diameter Obliquely. 


5 







6 


CONTENTS 


Chapter VII. Triangulation. 87 

Problem 65, Transition Piece, Square to Square—Problem 66, Reg¬ 
ister Box, Rectangular to Round—Problem 67, Register Box, Rec¬ 
tangular to Round, Vertical on Two Sides—Problem 68, Irregular 
Flaring Pipe Connection, Vertical on One Side—Problem 69, Tran¬ 
sition Piece, Round to Oblong—Problem 70, Twisted Rectangular 
Pipe—Problem 71, Transition Piece, Rectangular to Triangular— 
Problem 72, Irregular Fitting Forming a Transition from Round to 
Oblong, Upper Base Inclined 45°—Problem 73, Irregular T-Joint— 
Problem 74, Roof Collar, Square to Round, Base Obliquely In¬ 
clined—Problem 75, Scalene or Oblique Cone—Problem 76, Oblong 
Raised Cover with Semi-Circular Ends—Problem 77, Rectangular 
Raised Cover with Rounded Corners—Problem 78, Six-Pointed 
Star—Problem 79, Ash or Garbage Chute Head. 


Chapter VIII. Triangulation, Simplified Method.103 

Problem 80, Irregular Flaring Roof Collar—Problem 81, Roof Col¬ 
lar Having Round Top and Square Base—Problem 82, Three- 
Piece Offset Fitting—Problem 83, Three-Piece Reducing Elbow, 
Round to Round—Problem 84, Three-Piece Elbow, Oblong to 
Round—Problem 85, Three-Piece Transition Elbow, Round to 
Square—Problem 86, Furnace Boot, Round to Rectangular—Prob¬ 
lem 87, Two-Piece Tapering Y-Joint—Problem 88, Irregular Two- 
Branch Y-Joint—Problem 89, Two-Branch Y-Joint, Round to 


Round—Problem 90, Two-Branch Y-Joint, Round to Square— 
Problem 91, Irregular Two-Branch Fitting, Round to Round— 
Problem 92, Irregular Fitting, Round to Square—Problem 93, 
Three-Branch Fitting—Problem 94, Three-Way Branch—Problem 
95, Two-Way Transitional Branch Mitering with Horizontal and 
Vertical Connections—Problem 96, Irregular T-Joint, Rectangular 
to Round—Problem 97, Two-Way Elbow Fitting—Problem 98, 
Ship Ventilator with Round Mouth and Base—Problem 99, Ship 
Ventilator with Round Base and Elliptical Mouth. 

Chapter IX. Skylights. 136 

Problem 100, Flat Skylight—Problem 101, Double-Pitch Skylight 
—Problem 102, Hipped Skylight. 

Chapter X. Special Problems. 149 

Problem 103, Rectangular Flaring Pan with Folded Corners—Prob¬ 
lem 104, Round Ventilator with Molded Base—Problem 105, Ball, 
Development by Zones—Problem 106, Ball Development by Gores 
—Problem 107, Base for Chimney Top, Rectangular to Round— 
Problem 108, Chimney Hood—Problem 109, Florist’s Watering Pot 
—Problem 110, Automobile Measure—Problem 111, A Pieced El¬ 
bow Intersecting a Round Pipe—Problem 112, Round Ventilator 
with Square Base—Problem 113, A Round Pipe Intersecting an El¬ 
bow Miter—Problem 114, Two-Branch Duct Fitting, Square to 
Rectangular. 





















. 



























PART ONE 

Drafting Principles 


chapter i 

Drawing Equipment 


Equipment.—The following list comprises the equipment re¬ 
quired for a course in sheet-metal pattern drafting: Drawing 
board, 24" x 30"; T square., 30"; 45° triangle, 10"; 30° x 60° tri¬ 
angle, 10"; architect’s scale, 12"; 4H drawing pencil; pencil 
erasing rubber; thumb tacks; detail paper; also a set of drawing 
instruments consisting of the following pieces: 5" dividers, 5 \" 
compasses, 3" bow spacers, 3" bow pencil, ruling pen, bow pen 
and irregular curves. 

Paper.—The paper in general use for sheet-metal pattern 
drafting is known as brown detail paper. It can be bought of 
almost any width, in large or small rolls, and is sold by the yard 
or pound. The paper should be of medium thickness, very 
strong and tough, because a shop drawing is likely to be sub¬ 
jected to considerable rough usage. If a finished drawing is to 
be made, white drawing paper should be used. It can be ob¬ 
tained in almost every conceivable grade and a variety of sizes. 

Pencils.—For working drawings, full size details, etc., on 
manilla paper, a 4H pencil is quite satisfactory. For develop¬ 
ing miter patterns in which the greatest accuracy is required, a 
5H pencil is generally used. The accuracy of drawings depends, 
in a great measure, upon the manner in which the pencils are 


sharpened. To sharpen the pencil, remove the wood from both 
ends by means of a sharp knife, exposing about 3/8" of lead. 
One end should then be sharpened to a round point, and the other 
to a chisel point or a wedge-shaped end. This operation should 
be done with a fine file or pencil sharpener. A strip of No. 0 
sandpaper, 4" long and 3/4" wide, glued upon a thin strip of 
wood, will be found very serviceable. 

The chisel end is used for drawing straight lines, and the 
conical point for free-hand sketching and marking dimensions. 
A soft pencil should never be used for drawing, because it be¬ 
comes dull after drawing a few lines. This makes it impossible 
to draw fine, sharp lines and keep the paper clean. 

Preparation of the Paper.—The paper is fastened to the 
board by means of thumb tacks, and care must be taken to have 
it lie perfectly flat on the board, so that it will have no wrinkles. 
To do this, proceed as follows: Place the long edge of the paper 
parallel with the long edge of the board, the paper being within 
about three inches of the lower and left-hand edge of the board. 
Insert a thumb tack in the upper right-hand corner and press it 
in until it is flush with the surface of the paper. Next, place the 
left hand on the paper near the upper right-hand corner; then 


9 



10 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


slide the hand toward the lower left-hand corner, removing all 
wrinkles, and insert a thumb tack as before. Lay the left hand 
on the middle of the sheet and slide it toward the upper left- 
hand corner; holding it there, press in the third tack. Slide the 
hand from the center of the paper toward the lower right-hand 


corner, and insert the fourth tack, completing the operation. 
During damp weather, when conditions are such as to cause the 
paper to swell and become very wavy and loose, remove three of 
the tacks and again fasten the paper as before, squaring the sheet 
by the most important line on the drawing. 


i 



















CHAPTER II 

Practical Geometry 


In presenting this subject to the student, no attempt has been 
made to give a complete course in geometry. Problems have 
been selected which will be of the greatest assistance to the pat¬ 
tern draftsman, and which are composed of examples that are 
used in every-day practice. They are arranged in logical order, 
and teachers and students will find this selection to cover the 
subject properly with reference to order and extent. 

Practical geometry is the science of geometry adapted to prac¬ 
tical purposes by omitting theoretical demonstrations. Every 
student whose aim it is to become a proficient sheet-metal drafts¬ 
man should have a fair knowledge of the subject. The problems 
should be carefully studied and worked with great accuracy, as 
the technical skill acquired in the use of the drawing instruments 
will be of great value in later work. 

Geometrical Problems 

When the problems herein given have been carefully studied, 
draw each problem, completing each step in the construction 
before proceeding to the next. All lines should be as sharp and 
fine as consistent with clearness. 

In the geometrical figures, the given and required lines are 
shown in full heavy lines, and the construction in full light lines. 

Preparation of Plates.—The size of paper recommended for 
the problems of this course is 15" x 20". The size of each plate 
is to be 14" x 18", having a border line all around 1/2" from the 
edge of the plate, leaving the space inside of the border line 13" 
x 17". Divide the plate into two equal parts by means of a 


horizontal line. Using the scale, divide the length of plate into 
three equal parts, as shown by the vertical lines. This divides 
the drawing plate into six rectangular spaces. The problems 
should be drawn as near the center of each space as possible. 

Fig. 1. To bisect a straight line MN, or the arc of a 
circle MON.—Let MN 3-1/4 inches long be the given line 
which it is required to bisect. With centers M and N, and any 
radius greater than one-half of M N, describe the arcs 1 and 2. 
Through the points of intersection of these arcs draw a line, and 
the points of intersection with the given line M N, and the arc 
MON, shown by OA will give the required points. 

Fig. 2. To erect a perpendicular from a given line.— 

Draw the line AB about 3-1/4 inches long. Locate point C near 
the middle of line AB. With the point of the compass on A and 
any radius greater than AB, describe an arc at 1. On B, with the 
same radius, describe the arc 2. Through the intersection of 
arcs 1 and 2, draw the line EC, which will be the required per¬ 
pendicular. 

Fig. 3. To erect a perpendicular near the end of a given 
line.—Draw AB 3-3/4 inches long. About 1/2 inch from# lo¬ 
cate the point C, from which the perpendicular is to be erected. 
With C as center and with any convenient radius, describe the 
arc 1 -2. Using the same radius, step off this distance from 1 to 
3 and 3 to U- Using any radius with 3 and 1+ as centers, describe 
arcs 5 and 6 intersecting each other at 7. Draw a line from C 
thru 7, which will give the required perpendicular at the given 
point C. 


12 


PRACTICAL GEOMETRY 


13 























14 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Fig. 4. To erect a perpendicular at the end of a given 
line.—Draw AC, the given line, 3-3/4 inches long. Set the 
point of compass on A, and with any radius describe the arc B2. 
OnB with the radius AB, describe the arc 3 intersecting arc B2 at 
E. Thru E draw line BF indefinitely; with radius BE, describe 
arc U, intersecting line BF at M. Connect MA, which will be 
the perpendicular required. 

Fig. 5. To draw a line parallel to a given line.—Draw 
AB 3-1/2 inches long. Near the end of the line at 1, set the 
point of the compass, and with a 2-inch radius, describe arc 2. 
With the same radius on the point 3, describe the arc U- Then a 
line drawn touching the arcs 2 and U will be parallel to AB. 

Fig. 6. To draw a circle and its properties.—Draw AB 
3-3/4 inches long. Bisect AB at C. With point of compass at 
C, and radius CA, describe the circumference of the circle. The 
diameter of a circle is any straight line drawn thru the center to 
opposite points of the circumference as AB. The radius of a 
circle is any line as CA and DC, drawn from the center to any 
point in the circumference; two or more such lines are radii, the 
plural of radius. An arc of a circle is any part of the circum¬ 
ference as EG. The sector of a circle is the part of a circle in¬ 
cluded between the radii and the arc which they intercept, as 
ACD. A segment of a circle is a part cut off by a chord, as 
EFG. A chord of a circle is a straight line joining the extreme- 
ties of an arc, but not passing thru the center, as EF. 

Fig. 7. To draw a tangent from any given point on a 
circle.—With point of compass at A and radius AB, describe a 
circle 3-1/2 inches in diameter. Thru point B and center A draw 
a straight line. A perpendicular drawn thru point B will give 
the required tangent, as CD. 

Fig. 8. To bisect a given angle.—Draw the given angle 
ABC. With any convenient radius and B as center, describe the 


arcs 1 and 2. With the same or a larger radius and 1 and 2 as 
centers, describe arcs intersecting at 3. Draw a line from 3 
to B, which divides the angle ABC into two equal parts. This 
problem shows how to obtain the miter line between the two 
parts of an elbow or sheet-metal molding. 

Fig. 9. To draw an equilateral triangle, one side being 
given.—Draw AB 2-3/4 inches long. With A as center and 
AB as radius, describe arc 1. With B as center and the same 
radius, describe arc 2 intersecting the former arc at C. Draw 
the lines BC and AC, and ABC is the required equilateral tri¬ 
angle. 

Fig. 10. To construct an angle similar to a given angle. 

—Let ABC be the given angle. With A as center and with any 
radius, describe arc 1-2, touching both sides of the angle. Draw 
line EF equal to AB. With E as a center and radius A2 of the 
given angle, describe arc 3~U- With U as center and radius 1-2, 
describe arc 5 intersecting arc 3-U at G. A line drawn from E 
thru point G, completes the angle equal to ABC. 

Fig. 11. To draw a triangle equal to any given triangle. 
—Draw the given triangle ABC and line 1-2 equal to AB. With 
the radius AC and the center at 1, describe the arc 3. With the 
center at 2 and the radius BC, describe arc U intersecting arc 3 at 
5. Draw lines 1-5 and 2-5, which will give the required tri¬ 
angle equal to the given triangle ABC. 

Fig. 12. To construct an irregular angular figure simi¬ 
lar to a given figure.—Draw line AB 3-1/4 inches long and 
construct trapezium ABCD. To copy this figure in exactly the 
same size as it is given, draw line 1-2 equal in length to AB. 
With A as center and Ae as radius, describe arc ej. With the 
same radius and 1 as center, describe arc 3-U. With e as center, 
describe the arc g. With the same radius and 3 as center, de¬ 
scribe arc 5 intersecting arc U at 6. Draw a line from 1 thru 6, 


PRACTICAL GEOMETRY 


15 






















16 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


making 1-7 equal to AD. With# as center, describe arc mn. 
With the same radius and 2 as center, describe the arc 8-9. 
With m as center, describe arc h, cutting line .BC at o. With the 
same radius and 8 as center, describe arc 10 intersecting arc 9 at 
11. Draw line 2-12 thru point 11 , and make it equal in length 
to BC. Draw line 7-12 to complete the trapezium similar to the 
given trapezium ABCD. 

Fig. 13. To construct a square from a given side.— 

Draw line AB 3-1/2 inches, the length of the given side. With 
A as center and AB as radius, describe arc B-l indefinitely. 
With the same radius and B as center, describe the arc A-2, in¬ 
tersecting arc 1 at C. Bisect AC at D thru intersecting arcs at 
E. With C as center and radius CD, describe arcs 3 and 1+, inter¬ 
secting arcs 1 and 2 at F and G. To complete the square, con¬ 
nect AF, FG and GB. 

Fig. 14. To construct a regular pentagon in a given 
circle.—With A as center and with the compasses set to 1-7/8 
inches, describe the circle BCDE. Draw the two diameters BD 
and EC perpendicular to each other. Bisect the radius AB by 
the line passing thru AB at 1. With 1 as center and 1-E as ra¬ 
dius, describe the arc, locating point 2. With E as center and 
the distance E-2 as radius, describe an arc cutting the circum¬ 
ference of the circle at 3 and 1+. Using the same radius with 3 
and If. as centers, describe the arcs 5 and 6. Connect points 
E-1+-5-6-3, which completes the pentagon. 

Fig. 15. To construct a pentagon from a given side.— 
Let AB be the given side. With A as center and with the com¬ 
passes set to 1-3/4 inches, describe the semi-circle BEC. Divide 
BEC into five equal parts, and from A draw lines thru the divi¬ 
sions 1-2-E. With AB as radius and E as center, describe the 
arc 3. With the same radius and B as center, describe the arc 1+. 
Draw the lines E-3, 3-1+ and 1+-B to complete the figure. 


Fig. 16. To construct a hexagon from a given side.— 

Describe a circle with the radius AB 1-7/8 inches, which will be 
the length of the given side. Draw the diameter BC. With the 
radius AB and the centers C-B, describe the arcs 1-2-3-1+. Con¬ 
nect by straight lines C-l, 1-2, 2-B, B-3, 3-1+ and 1+-C, which 
completes the required hexagon. 

Fig. 17. To inscribe an octagon within a given circle.— 
With A as center, with the compasses set to 1-7/8 inches, de¬ 
scribe the circle 1-2-3-1+-5-6-7-8. Let this be the given circle 
in which it is required to inscribe a regular octagon. Thru the 
center, draw lines BC and DE perpendicular to each other, cut¬ 
ting the circumference of the circle at 1-5 and 7-3. Bisect the 
angles DAB and DAC, and let the bisector of each angle meet 
the circumference at 2 and 8. Draw the diameters 8-1+ and 
2-6. Straight lines drawn from 1-2, 2-3, 3-J+, etc., will form 
the required octagon. 

Fig. 18. To inscribe an octagon within a given square. 

—Draw line AB 4 inches long, and construct the given square 
ABCD. Draw diagonal lines DB and AC. With B as center 
and BG as radius, describe the arc 1-2. With the same radius 
and ADC as centers, describe arcs 3-1+, 5-6, 7-8. Straight lines 
drawn from 6-2, 2-8, .8-3, etc., will complete the required octa¬ 
gon. 

Fig. 19. To construct an octagon, one side being given. 

—Draw line AB 1-1/4 inches long, which is the length of the 
given side. Extend AB indefinitely, as shown by 1 and 2. 
From A and B erect indefinite perpendiculars as AC and BD. 
With A andB as centers, using any radius, draw the arcs 1-3 and 
i+-2. Bisect the angles l-A-3 and 1+-B-2 by 5-A and B-6. On 
these two lines set off A-7 and B-l2, equal to AB. From 7 and 
12 erect the perpendiculars 7-8 and 12-11, equal to AB. With 
8 and 11 as centers and AB as radius, describe arcs 9-10, inter- 


PRACTICAL GEOMETRY 


17 










































18 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


secting perpendiculars AC and BD at 9 and 10. Connect 8-9, 
9-10 and 10-11, which completes the required octagon. 

Fig. 20. To draw a circle thru any three given points 
not in a straight line.—Let ABC be the given points not in a 
straight line. Draw the lines CA and AB. Bisect the' line CA 
by EF, as shown. Also bisect AB by the line GF, and the inter¬ 
section of the bisecting lines at F will be the center of the re¬ 
quired circle. Then with F as center and FB as radius, describe 
the circumference thru points ABC. 

Fig. 21. To find the center of a circle when the circum¬ 
ference is given.—Let ABC be the given circle. From any 
point on the circle as B, with any radius, describe the arc 1-2. 
Then from the points A and C, with the same radius, describe 
the intersecting arcs 3-4 and 5-6. Thru the points of intersec¬ 
tion draw the lines 7-8 and 9-10, which will meet in x. Then x 
will be the center of the circle. 

Fig. 22. To describe the segment of a circle of any 
given chord and height.—Draw the line AB 3-3/4 inches long, 
which will be the given chord. Draw the perpendicular mn in¬ 
definitely, and make Pm the given height 1-1/8 inches long. 
Connect mB and bisect mB by the line CG, intersecting the per¬ 
pendicular mn at C. Then C will be the center from which to 
describe the segment AmB. 

Fig. 23. To inscribe an equilateral triangle in a circle. 

—Draw the line AB 3-3/4 inches long, which will be the diameter 
of the given circle. With B as center and BC the radius of the 
circle as radius, describe the arc FCG. To complete the in¬ 
scribed triangle, connect by straight lines FA, AG and GF. 

Fig. 24. To inscribe a circle in a given triangle.—Draw 
the line AB 3-3/4 inches long. Make AC 3 inches, and CB 4 
inches in length. Bisect the angles CAB and ACB. The inter¬ 
section of the bisectors at m will be the center of the circle 


which can be described, touching all three sides of the triangle. 
The sides AB, BC and CA will be tangent to this circle. 

Fig. 25. To draw an ellipse when the diameters are 
given,without using centers.—Drawthe line 1-A 3-1/2inches 
long, which will be the required length. Bisect 1-A at C. Thru 
C draw DE 2-1/4 inches long, the required width. With C as 
center and C-l and CE as radii, describe the outer and inner 
circles, respectively, as shown. Divide one-quarter of the outer 
circle into any convenient number of parts; in this case, into 
five, as shown by l-2-3-l>-5-6. Divide the one-quarter inner 
circle into the same number, as shown from 1 to 6. From the 
points on the smaller circle, draw horizontal lines, and thru the 
points on the larger circles, draw vertical lines. The points a, b, 
c, d, where the horizontal and vertical lines intersect, are points 
on the ellipse. Using an irregular curve, trace a line thru the 
points thus obtained, completing one-quarter of the ellipse. 

Fig. 26. To draw an approximate ellipse when length 
and width are given, using circular arcs.—Draw the line 
AB 3-1/4 inches long. Bisect AB at m, and draw the width CD 
2-1/8 inches long. On the length AB, set off the width CD from 
B to 3, and divide the balance 3A into three equal parts, as 
shown by 1,2, 3. With m as center and a radius equal to the 
length of two of these parts, describe arcs cutting AB in E andE. 
With EE as radius andE andE as centers, intersect arcs at 4 and 
5. Draw lines from 4 thru E and F as 4 -6 and 4 -7, and lines 
from 5 thru EF as 5-8 and 5-9. With 4 and 5 as centers and 5-C 
and 4 -D as radii, describe the arcs GDH and OCP. WithE and 
F as centers and radii equal to EA and FB, describe the arcs 
GAO and PBH, completing the ellipse. 

Fig. 27. To draw an approximate ellipse, the major 
and minor axes being given.—For many purposes in sheet- 
metal drawing, it is sufficiently accurate to describe the ellipse 


PRACTICAL GEOMETRY 


19 

































20 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


by means of circular arcs, and where centers must be used in de¬ 
veloping patterns for flaring articles. Draw the major diameter 
AB 3-7/8 inches long, and the minor diameter CD 3 inches in 
length. On the line CD lay off mF and mG, equal to the differ¬ 
ence between the major and minor diameters. On the line AB 
lay off mE, and mH equal to three-quarters of mG. Connect 
points FHGE, and extend the lines. With center E and radius 
EA, describe arc RAO. With center F and radius FD, describe 
arc ODP. In a similar manner, describe arcs PBS and SCR 
from centers G and H. This is not a practical method when the 
major diameter is more than twice the minor. 

Fig. 28. To draw an ellipse by intersection of lines.— 
Draw the major axis AB 3-1/2 inches long, and the minor axis 
mA' 2-1/4 inches. Thru m parallel to line AB draw line CD. 
From points A and B erect perpendiculars to line CD. Divide 
lines AC and DB into any number of equal parts; in this case, 
four, and draw lines from points 1,2, 3, etc., to m. Divide An 
and nB into the same number of equal parts, and draw lines 
from A' thru these points intersecting the similarly numbered 
lines drawn from the points on the line CA and DB. Thru these 
points of intersection, trace the semi-ellipse AMB. 

Fig. 29. To draw an egg-shaped oval with arcs of cir¬ 
cles.—With a radius of 1-3/8 inches and C as center, describe 
the circle AmBn. Thru the center C, perpendicular to AB, 
draw the line mn. Thru n draw Bn and An indefinitely. On A 
and B as centers, with AB as radius, describe the arcs B H and 
AG. With n as center, describe the arc GH to complete the 
figure. 

Fig. 30. To draw an ellipse by means of a pencil and 
thread.—Draw AB, the major axis, 3-3/4 inches long. Bisect 
AB at m. Thru m draw the perpendicular CD 2-1/2 inches 
long. Take Am one-half the length of the major axis for radius, 


and with C as center, describe the arc GH. Drive pins at C, G 
and H; then tightly tie a thread around the three pins CG H. 
Remove the pin at C, and, placing a pencil at this point, keeping 
the thread tightly stretched, describe the ellipse. 

Fig. 31. To draw a parabola, having given the axis AB 
and the double ordinate FD.—Draw AB 3-1/2 inches long, 
and FD perpendicular to AB, 4 inches long. Draw EF and CD 
parallel and equal to AB. Divide EF and BF into the same 
number of equal parts. From the divisions on BF, draw lines 
parallel to the axis AB, and from the divisions on EF, draw lines 
to the vertex A. The points of intersection of these lines 1 and 
1 , 2 and 2, etc., are points on the required curve thru which it 
may be traced. In like manner, obtain the opposite side. 

Fig. 32. To draw a hyperbola, the axis, a double ordi¬ 
nate and its distance from the vertex being given.—Draw 
the double ordinate FD 3-3/4 inches long. Perpendicular to 

FD, draw the axis BA 3-3/8 inches long. On the line AB locate 
M 1-3/8 inches from the vertex A. Thru M draw EC perpen¬ 
dicular to AB; then draw FE and DC perpendicular to FD, in¬ 
tersecting FE and DC in E and C. Divide FE and DC into the 
same number of equal parts, and from points 1, 2, 3, etc., on 

FE, draw lines to the vertex M. From points on FB, draw lines 
to the vertex A. The intersection of these lines 1 and 1,2 and 2, 
etc., will be points in the required hyperbola. 

Fig. 33. To draw an equable spiral.—Draw the line A6 
4-5/8 inches long. Bisect A6 at 0, and with 0 as center and OA 
as radius, describe the circle A3-6-9. Divide the circle into twelve 
equal parts, and to the points on the circumference draw radial 
lines from the center at 0. Divide AO into as many equal parts 
as the spiral is to have revolutions; in this case, two. Divide 
each space into twelve equal parts, the same number of parts as 
there are divisions in the circle. With 0 as center and 0-1,0-2 


PRACTICAL GEOMETRY 


21 












































22 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


0-3, 0-4, etc.,as radii, describe arcs intersecting similarly num¬ 
bered radial lines, as shown. Thru the points of intersection 
thus obtained, trace a curved line, completing the required 
spiral. 

Fig. 34. To draw a scroll to a specified width.—This 
problem shows how to obtain the pattern for a simple scroll 
much used in the construction of brackets and modillion blocks; 
also as a terminal finish in sheet-metal cornice construction. 
Draw AB the given width, 4-1/4 inches long. On the line AB 
locate point C one inch from A, for the width of the band. Make 
AD equal to one-third of AB. Bisect AB, obtaining the point 
m. Now, take one-eighth of the distance AD and set it off be¬ 
low point m, thus obtaining the point G. From G draw a hori¬ 
zontal line, and using the 45° triangle, draw from A a line inter¬ 
secting the horizontal line at F. Draw the rectangle GFEB. 
From E draw a line, making an angle of 45° with BE, intersect¬ 
ing GF at H. Draw the square GHOP, and divide GP into 
four equal parts; on two of these parts, construct the inner square 
1-2-3-1+ . The arcs forming the outline of the scroll are de¬ 
scribed in the following manner: With G as center and radii 
GC and GA, describe the arcs AF and C-6. From H as center, 


with radius H-6 and HF, describe the arcs FB and 6-7. Con¬ 
tinue the operation by describing arcs tangent to those first 
drawn, using points 0, P, and 1,2,3, 4 at the angles of the smaller 
square as centers for describing the successive quadrants. 

Fig. 35. The Helix.—The Helix is a curve formed by a 
point moving around a cylinder and at the same time advancing 
along the line of its axis a fixed distance for each revolution* 
The distance advanced at one revolution is called the pitch. 
The line described upon the surface of the cylinder is repre¬ 
sented by a flexible cord wound around the cylinder; it is shown 
in actual practice by the thread of a screw. Let the circle 1-5- 
9-13, having a diameter of 2-3/4 inches, represent a plan view of 
the cylinder. Draw the elevation ABC-1, and make C-l, the pitch 
of the helix, 3-1/2 inches long. Divide C-l into a number of 
equal parts; in this case, sixteen, and divide the circle into the 
same number of equal parts, beginning at point 1, as shown in 
the drawing. From the points on the circle, as 1,2, 3, 4, etc., 
draw vertical lines, intersecting like numbered horizontal lines 
drawn from similarly numbered points on the pitch line 1-C. 
Thru the points thus obtained, trace the helical curve, making 
one revolution. 


PRACTICAL GEOMETRY 


23 






































































CHAPTER III 

Practical Pattern Drafting 


Sheet-metal pattern drafting is founded upon those principles 
of geometry which relate to the surfaces of solids, and may be 
described as the development of surfaces. Sheet-metal articles 
are hollow, and are considered in the process of pattern drafting 
as though they were the coverings of solids of the same shape. 

The different methods for developing the patterns for forms 
with which the pattern draftsman has to deal may be divided 
for convenience of description, into three general divisions: 

First, Parallel Line Developments, which is used in developing 
patterns for moldings, pipes, elbows and regular continuous 
forms, or may be called parallel forms. 

Second, Radial Line Developments. This method is used in 
developing patterns for regular tapering forms by means of 
radial lines, converging to a common center. The forms having 
for their base the circle, or any of the regular geometric figures 
which terminate in an apex. 

Third, Triangulation. This method is used in developing 
patterns for irregular forms which cannot be developed by either 
the parallel or radial-line methods. 

All of the problems that will follow should be carefully stud¬ 
ied, drawn on detail paper, and the drawings proved by paper 
models when it is not possible to construct the problems from 


sheet metal. These models will at once show any error in the 
drawings which might otherwise be overlooked. 

Practical work-shop problems, such as arise in every-day 
practice, are presented; an actual trade object forms the subject 
of each problem. No work is introduced that is not practical 
and likely to arise in the every-day work of a sheet-metal pattern 
draftsman. The problems should be taken in regular order, 
and drawings constructed as called for by the text. 

Some of the problems are fully developed, and the demonstra¬ 
tions are made more explicit than others. Less important de¬ 
tails of the work are sometimes omitted and certain parts of the 
operation are described in a general way upon the supposition 
that the problems in which the development is fully described 
would naturally be studied first. The problems should be 
drawn according to the dimensions given, and all problems will 
then be of ample size to permit each step in the drawing of the 
pattern to be clearly shown, and of proper size for constructing 
from sheet metal. Allowances for seams, joints, etc., are shown 
in some of the problems, but if a joint is required at a place other 
than where shown, it can be changed at the discretion of the 
draftsman without changing the principles involved in the de¬ 
velopment. 


24 



PART TWO 

Parallel Line Developments 

chapter IV 

Pattern Problems 


Under this head, problems are presented in which the patterns 
are developed by means of parallel lines. This method is used 
in developing the pattern for any form the opposite lines of 
which are parallel, such as elbows, T-joints, roof gutters, cor¬ 
nices, skylights, etc.; also in patterns for miters that occur in 
joining moldings, pipes and all regular continuous forms at any 
angle and against any other form or surface. 

There are certain fixed principles that apply to developments 
by this method, and the following rules should be carefully ob¬ 
served by the student and draftsman. 


1) A plan and elevation must first be drawn, showing the 
article in a right position, in which the parallel lines of the solid 
are shown in their true length. 

2) The pattern is always obtained from a right view of the 
article in which the miter line or line of intersection is shown. 

3) A stretchout, or girth, line is always drawn at right angles 
to the parallel lines of the solid, upon which is placed each space 
contained in the section or plan view. 

4) Measuring lines are always drawn at right angles to the 
stretchout line of the pattern. 


25 





26 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


5) Lines drawn from the points of intersection on the miter 
line in the rightview, intersecting similarly numbered measuring 
lines drawn from the stretchout, will give points showing the 
outline of the development. 

6) A line traced thru the points thus obtained will give the 
desired pattern. 

Problem 1. Pattern for Two-Piece Elbow in 
Round Pipe 

In Figure 36, let AB-l-CD-7 be the elevation of a two-piece 
90° elbow. First, draw 
the elevation. Then, be¬ 
low the elevation, de¬ 
scribe a circle represent¬ 
ing the profile or plan, 
shown at F. As each 
half of the pattern is 
symmetrical, draw a line 
thru the planF, and di¬ 
vide the upper half of 
the circle into a number 
of equal parts, as shown 
from 1 to 7. From 
these points perpendic¬ 
ular lines are drawn, in¬ 
tersecting the miter line I -7 as indicated. Then, at right angles to 
the vertical arm of the elbow D-7, draw the stretchout line FG, 
and upon this line step off twice the number of spaces shown 
in the plan, which will give the circumference of the elbow. 
From these points and at a right angle to FG, draw measuring 
lines which are intersected by like numbered lines drawn at a 
right angle to the cylinder from similarly numbered points on 
the miter line 1-7 in the elevation. A line traced thru points 


thus obtained will be the pattern for the vertical arm of the 
elbow, as shown by FGLHJ. 

The irregular curve traced thru the points of the pattern is the 
only one required for both pieces of the elbow, and to save mate¬ 
rial, the pattern for the upper arm of the elbow is generally ob¬ 
tained in the following manner: 

The stretchout of both pieces being of equal length, extend the 
outer lines of the pattern to M and R, as shown in the drawing, 
and make JM and RL equal in length to A-7 in the elevation. 
Draw a line from M to R; then JLRM will be the pattern for the 
upper arm of the elbow, having the seam at A-7 on the outside 
or heel of the elbow. The seam on the lower arm is on the inner 
side or throat, as shown by C-l in the elevation. 

This method of development is applicable to any pieced elbow, 
no matter what the shape of the pipe may be or the angle re¬ 
quired. 

Problem 2. Conductor Elbow 

In Figure 37, let ABFRG be the elevation of a two-pieced el¬ 
bow, the circle representing the profile or plan being three inches 
in diameter. Draw the elevation and make BEF an angle of 
120°. Bisect the angle BEF and draw the miter line EG. Divide 
the upper half of the profile into a number of equal parts, and 
from these points draw vertical lines intersecting the miter line 
EG in the elevation. 

The stretchout line as shown by the line ab is next drawn at 
right angles to the lower arm of the elbow, and the patterns for 
both pieces are developed in the same manner as the 90 ° elbow, 
which is fully explained in Problem 1. 

Problem 3. Rectangular Conductor Shoe 

The principle here involved and the method of procedure are 
exactly the same as in Problem 2, the only difference being in the 



Problem 1. Two-Piece Elbow, 
Round Pipe. 




PATTERN PROBLEMS 


27 



0 ) 



















































































28 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


shape of the pipe. In Figure 38, let 1-2-3-^-x in the plan 
represent a rectangular pipe having the seam at x. Draw the 
plan and elevation according to the dimensions given on the 
drawing, and find the miter line by bisecting the angle in the 
usual manner. 

Draw the stretchout line x-x at right angles to the vertical 

arm A, and on this line step 
off the spaces x-1,1 -2,2-3, 
3-1* and b-x of the plan C. 
From the points thus ob¬ 
tained draw the usual 
measuring lines, which are 
intersected by horizontal 
lines drawn from points 
2-3 and 1-x-U on the miter 
line in elevation. Lines 
drawn connecting these 
points will give the desired 
pattern. 

The small circles marked 

near the extremity of each 
Problem 3. Rectangular Conductor Shoe. measuring ] ine on the pat- 

tern indicate that the metal is to be bent along this line when 
forming the work into the required shape. 

Problem 4. Two-Piece Elbow in an Oblong Pipe. Case I 

The only difference to be observed in developing patterns for 
elbows in oblong pipes, as compared with the same operations in 
connection with round pipes, lies with the profile or plan. 

The plan is to be placed in the same position as shown in the 
development of patterns for elbows in round pipes, but is placed 
with the long or narrow side to the view, as the requirements of 
the case may be. 


In Figure 39 is shown the elevation, plan and pattern for a 
two-piece oblong elbow having semi-circular ends. The narrow 
side of the oblong is presented to view, with the seam in the cen¬ 
ter of the long side, as shown by x in the plan. Draw the plan 
and elevation, and divide the upper semi-circular end into a 
number of equal parts. From these parts draw vertical lines to 
the miter line CF. At right angles to the vertical arm EF, draw 
the stretchout line x-x, and thru the points in it draw the usual 
measuring lines. From the points on the miter line CF, draw 
horizontal lines, intersecting similarly numbered measuring lines, 
and a line traced thru these points will give the required pattern. 

Problem 4. Case II 

In Figure 40 is shown the plan and elevation of a right-angled, 

two-piece elbow in an oblong 
pipe, having the long side of 
the oblong presented to view, 
as shown by the position of 
the plan. 

Draw the plan and eleva¬ 
tion and develop the pattern 
for the vertical arm EF. The 
order of procedure is the 
same as that given for the 
drawing in Case I, although 
the results in the two cases 
are different in consequence 
of the position of the profiles. 
The drawing should present 
no difficulties to the attentive student. 

Problem 5. Pipe and Roof Flange 

Figure 41 shows the method of developing the patterns for a 








PATTERN PROBLEMS 


29 

































































30 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


pipe and roof flange used by plumbers and sheet-metal workers 
when flashing around vent pipes and stacks that come thru the 
slanting sides of a roof. 

Drawthe roof lineAB at an angle of 45 °. Then draw a side view 
of the pipe C, and immediately above it and in line with the pipe, 
draw the profile F. Divide the upper half of the profile into a 
number of equal parts, and from these points draw vertical lines 
to the line AB, representing the pitch of the roof. To develop 

the pattern for the pipe C, 
draw the stretchout line ab 
at right angles to the ver¬ 
tical side of the pipe, arid 
obtain the pattern in a 
manner similar to the de¬ 
velopment of the lower 
arm of the two-piece elbow 
shown in Figure 36. 

The pattern for the 
opening in the roof flange 
is shown at G, and is de¬ 
veloped in the following 
manner: First, draw lines 
at right angles to the roof 
line AB from the points 1 
to 7. Then, at right angles 
to these lines draw the line 
J K thru the center of the 
roof plate efgh. On the line J K place one-half of section F, as 
shown by M, and divide the half circle into the same number of 
equal spaces to correspond to section F. From these points in 
M draw lines parallel to J K, intersecting similarly numbered 
lines that have been drawn from the points on the line AB. A 


line traced thru the points thus obtained will be the pattern for 
the opening in the roof plate. 

Problem 6. Pipe and Roof Flange for Ridge of Roof 

In Figure 42, let ABC be the length of the roof plate and a sec¬ 
tion of the roof against which the flange is to fit. Draw an ele¬ 
vation of the pipe 
G, and the profile 
H in their proper 
position. Then de¬ 
velop patterns for 
the pipe and the 
opening in the roof 
flange in the same 
manner as de¬ 
scribed in Problem 
5. Since both 
halves of the open¬ 
ing in the roof 
flange from the 
point A are the 
same, both halves 
of the pattern may 
be obtained at one operation, the line BA may be extended 
across the pipe to m and used in place of AC. When the roof 
line is extended in this manner, it will be seen that the method 
of developing the pattern for the opening is identical with that 
shown at G in Figure 41. 

Problem 7. An Octagonal Pipe Fitting Over the Ridge 
of a Roof 

Applying the method given in Figure 42, develop the pattern 
for the octagon shaft shown in Figure 43. Let F be the section 

























PATTERN PROBLEMS 


31 


m 



© 






























































































































32 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


and E the elevation of an octagon pipe mitering against a 
double-pitched roof represented by the lines BA and AC. Draw 
a vertical line from the point A to the section F, locating the 
points a and b which are placed in their proper position on the 
stretchout line. 

From these points draw measuring lines which are intersected 
by a line drawn from the point A, representing the ridge of the 
roof, as shown in the pattern G. 

Problem 8. T-Joint Between Pipes of Same Diameter 

Figure 44 shows the method of developing the patterns for 

two cylinders of the same 
diameter intersecting at 
right angles. 

Draw the elevation and 
place the horizontal pipe A 
in the center of the vertical 
pipe B. Then draw the half 
section of the horizontal 
pipe A and divide it into a 
number of equal parts, as 
shown from I to 7 in D. 
Draw the half section G of 
the vertical pipe and divide 
it into the same number of 
parts as section D, placing 
the numbers in their proper 
position as shown. In the 
half section D the points 1 and 7 are on the top and bottom, 
.while the point 4 is on the long side of the pipe. As both pipes 
are the same diameter, both halves of pipe A will miter with 
one-half of pipe B, and when looking down upon the end of the 


vertical pipe, point 4 will intersect the vertical pipe on the side, 
as shown by point 4 in half section G. 

Horizontal lines are now drawn from the points in section D, 
which are intersected by vertical lines drawn from similarly num¬ 
bered points in section G. 
Lines drawn thru these points 
of intersection will give the 
miter line. The two pipes 
being of the same diameter, 
the miter line is represented 
by straight lines at an angle 
of 45°, shown by abc. To 
obtain the pattern E for the 
horizontal pipe A, draw the 
stretchout line mn, upon 
which step off twice the 
number of spaces contained 
in the half section D. From 
these points draw the usual 
measuring lines which inter¬ 
sect by vertical lines drawn from similarly numbered points 
on the miter line abc. 

A line traced thru these intersections will give the required 
pattern. The pattern F for the vertical pipe B is simply a rec¬ 
tangle in form, the length being equal to the circumference of 
the pipe, and the width being equal to the height. 

The pattern for the opening F is obtained in the following 
manner: Locate point 1 upon the upper edge of pattern C, 
which will be the center of the opening. On each side of point 1 
step off the spaces shown from 1 to 4 in the half section G, which 
will give the length of the opening, and from these points draw 
vertical lines which are intersected by horizontal lines drawn 









PATTERN PROBLEMS 


33 


from similarly numbered points on the miter line abc in the 
elevation. A line traced thru the points thus obtained will give 
the pattern for the desired opening. 

Problem 9. T-Joint Between Pipes of Different Diameters 

Figure 45 shows the plan and elevation of a T-joint, the pipes 
being of different diameters, the horizontal pipe B being placed 
in the center of the vertical pipe A at an angle of 90°. 

First, draw the plan and elevation, as shown in the drawing. 
After the outline of the 
small pipe H has been 
drawn in the plan view, 
draw the half section C 
and divide it into a 
number of equal parts. 

Then draw horizontal 
lines from these points, 
intersecting the large 
pipe G, as shown. Next, 
draw the half section F 
on the end of the small 
pipe in the elevation, 
and divide it into the 
same number of spaces, 
as the half circle is a 
duplicate of the half 
section C in the plan. 

From the points in section F draw horizontal lines which are 
intersected by vertical lines drawn from similarly numbered 
points on the large circle G in the plan. A curved line traced 
thru these points of intersection will give the miter line between 
the two pipes. Develop the pattern for the small pipe and 


the opening in the large pipe in the same manner as explained 
in the previous problem, shown in Figure 44. 

The stretchout of the opening in the large pipe is shown by the 
figures 1-k-l in the plan. The spaces being unequal in length, 
they must be transferred separately to the stretchout line of the 
pattern. 

Problem 10. T-Joint Between Pipes of Same Diameter 
at an Angle 

Figure 46 shows the intersection of two cylinders of equal 
diameter at an angle of 45°. Let A represent the plan of the 
vertical pipe, and B its elevation. Draw the outline of the 
oblique pipe C at its required angle, and place the section D in 
its position, as shown. Space one-half of plan A, and section 
D into the same number of equal parts. Draw lines from 
these points intersecting in the elevation. A line drawn thru 
the intersections obtained in this manner will give the miter 
line between the two pipes, shown by abc in the elevation. Pat¬ 
tern E for the inclined pipe and pattern G for the vertical pipe 
are shown fully developed. 

The principles in this problem do not differ from those given 
in Figure 44. The problems are the same except in the position 
of the oblique pipe C, and the same principles are applicable, no 
matter what diameter the pipes may have, or at what angle they 
are joined. 

Problem 11. Y-Joint 

Figure 47 shows the elevation, partial development and di¬ 
mensions of a Y-joint, the diameter of each branch being the 
same. 

Draw the elevation, making the arms AB at an angle of 90 °. 

The miter line ba is obtained by bisecting the angle dac by the 
line bg. The pipes being of the same diameter, a half section of 
arm A shown at D is all that is required to obtain the points on 



Problem 10. T-Joint, Pipes of Same 
Diameter at an Angle. 







34 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


the miter lines. Divide the half section D into a number of 
equal parts, being careful to place a point on the quarter circle, 
as shown by the point U- Draw lines from these points to the 
miter lines bf and ba, as shown. Then draw the stretchout lines 
mn and develop patterns for the arms A and C, placing the 
seams on the short side of both pieces. 

Problem 12. Two Square Pipes of the Same Size Inter¬ 
secting at an Angle 

In Figure 48 is shown the intersection between two square 

pipes of equal size, the 
inclined pipe G being 
placed in elevation at 
an angle of 30° to the 
base line. 

Let ABCD represent 
a plan of the vertical 
pipe placed diagonal¬ 
ly, as shown, above 
which draw the eleva¬ 
tion E. In its proper 
position, draw the out¬ 
line and profile of the 
inclined pipe in both 
the plan and eleva¬ 
tion, numbering the 
corners 1-2-3-U, as shown by F and K. 

From the corner A in the plan, draw a horizontal line thru the 
profile F, locating the points a and b, as shown. These points 
must also be placed in their proper position in profile K, and on 
the stretchout line when developing the pattern for the pipe G. 
Develop the patterns for the vertical pipe E and the inclined 


pipe G; also the pattern for the opening in the vertical pipe. 
The principles explained in Figure 46 for the intersection of 
round pipes are also applicable for pipes that are square, rectan¬ 
gular, elliptical and oblong in form. 

Problem 13. Two Round Pipes of Unequal Diameters 
That Intersect Irregularly 

Figure 49 shows the elevation, plan and patterns of two cylin¬ 
ders of different diam¬ 
eters intersecting at an 
angle of 45 °. The posi¬ 
tion of the two pipes is 
such that the outline 
7-7 of the smaller pipe 
in the plan view is tan¬ 
gent to the circle that 
represents the large 
pipe. First, draw the 
plan A of the large pipe 
and the profile B of the 
smaller pipe in their 
proper position. 

Space the profile B 
into a number of equal 
parts, and from these 
points draw lines inter¬ 
secting the large pipe in the plan at A. 

Draw the elevation of the two pipes, as shown atF and H, and 
describe the circle that represents the profile of the small pipe, 
as shown at C. Divide the profile C into the same number of 
equal spaces as the profile B, and from these points draw lines 
parallel to the inclined arm indefinitely. From the points in 



Problem 12. Square Pipes of Same Diameter 
Intersecting at an Angle. 



Problem 13. Two Round Pipes of Unequal 
Diameters Intersecting Irregularly. 














PATTERN PROBLEMS 


35 




® 
































































36 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


circle A of the plan, draw vertical lines, intersecting the oblique 
lines drawn from points in profile C. A line drawn thru these 
points of intersection will be the miter between the two pipes. 
Note the position of the points on the circle B that represents 
the profile of the smaller pipe in the plan at 1,2,3, etc., and their 
corresponding location 
at 1,2, 3, etc., in the 
profile C in the eleva¬ 
tion. 

For the pattern of the 
small pipe, proceed as 
follows: Draw the 
stretchout line ab at 
right angles to the arm 

H, and draw the meas¬ 
uring lines, as shown, 
numbering them to cor¬ 
respond to the spaces in 
profile C. The seam has 
been placed on the short 
side of the pipe at point 

I. Therefore, com¬ 
mence numbering the stretchout line with 1 . From the points 
1, 2, 3, etc., on the miter line, draw lines at right angles to the 
arm H, intersecting similarly numbered measuring lines. 

A line traced thru these points will give the required pattern 
shown at G. 

The development of the large cylinder is shown at K. Draw 
the stretchout line and make x'-x' equal in length to the circum¬ 
ference of the circle shown at A in the plan. 

Reference to the plan shows that the small pipe intersects the 
surface of the larger from points 1 to 7 on the large circle, which 


will be the stretchout for the pattern of the opening in the large 
pipe. For the pattern of the opening, locate the point 7 on the 
stretchout line x’-x', the distance from x' to 7 being one-quarter 
of the circumference of the large pipe. Then take the divisions 
7-6,6-5,5-U, etc., on the large circle in the plan, and place them 
on the stretchout line x'-x', as shown. 

From these points draw vertical lines which are intersected by 
horizontal lines drawn from similarly numbered points in the 
elevation. 

A line traced thru these points of intersection will give the 
irregular outline of the pattern, shown at P in the drawing. 

Problem 14. An Octagonal Pipe Intersecting a Square 
Pipe Obliquely 

Figure 50 shows the plan, elevation and intersections of a 
square and octagonal pipe, the octagonal pipe being placed to 
one side of the center of the square pipe and inclined at an angle 
of 30° to the base line. Draw the plan of the square pipe, as 
shown at ABCD . Locate the point 3~U one-half inch from the 
cornerB, and draw the profile of the octagon pipe in the position 
shown at E. 

Draw the elevation and profile F, as shown, placing the num¬ 
bers on the profile in their proper position. It is necessary to 
use extreme care in numbering the points on the two profiles, in 
order that the position of the points on the profiles may be lo¬ 
cated in the same corresponding position with regard to one an¬ 
other. From the center of the square pipe shown at A in the 
plan, draw a line intersecting the profile E at x and y. 

The points x and y must be placed in their proper position in 
the profile F, also on the miter line and stretchout line, before 
developing the patterns. This problem introduces no new ele¬ 
ment in the development of solids by parallel lines. The prin- 



Problem 14. Octagonal Pipe Intersecting 
a Square Pipe. 





PATTERN PROBLEMS 


37 



© 

































































38 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


ciples are the same as those governing the development of the 
round pipes in the previous problem. Complete the problem 
by drawing the miter lines, and develop the patterns for the 
square and octagon pipes. Also a pattern for the opening in the 
square pipe. 


Problem 15. A Rectangular Pipe Intersecting a Round 
Pipe Obliquely 


In Figure 51 is shown the 
plan and elevation of a rec¬ 
tangular pipe intersecting a 
cylinder obliquely. Draw the 
plan and elevation, placing 
the oblique pipe C in the ele¬ 
vation at an angle of 45° to 
the base line. 

Draw the profiles of the 
rectangular pipe in the posi¬ 
tion shown at F and G. 

Develop the patterns for 
the rectangular pipe C, and 
the opening in the round 
pipe A, as shown at B, in ac¬ 
cordance with principles already explained. 



Problem 15. Rectangular Pipe 
Intersecting a Round Pipe. 


Problem 16. Four-Piece 90° Elbow 


Figure 52 shows the method of obtaining the patterns for a 
four-piece 90 ° elbow having a diameter of 5 inches; the length of 
the radius for the inner curve of the elbow being 3 inches. 

Draw the right angle ABC, and on the line BC lay off a dis¬ 
tance of 3 inches from B to n. With B as center and Bn as ra¬ 
dius, describe the quarter circle mn, which gives the required 
curve for the throat. 


Make nc the diameter of the elbow, and with BC as radius and 
B as center, describe the outer arc CA. 

The miter lines of the elbow, shown by EFG, are obtained by 
dividing the outer quarter circle AC into equal parts one less in 
number than the pieces required in the elbow; in this case, into 
three parts, shown by Ae, ee and ec. Bisect each of these spaces, 
and lines drawn from these points to the center B will give the 
joint or miter lines of the elbow. This method can be used to 
obtain the miter line for elbows at any angle having any desired 
number of pieces. Next, draw the half section D, and divide 

it into a number of 
equal spaces, and from 
these points draw ver¬ 
tical lines intersecting 
the miter line BG in 
the elevation. 

To develop the pat¬ 
tern for the first sec¬ 
tion of the elbow, draw 
the stretchout line 
HR, upon which place 
twice the number of 
spaces contained in 
half section D. From 
these points draw the 
usual measuring lines, 
which intersect lines 
drawn from similarly 
numbered points on the miter line BG. Thru the points thus 
obtained, trace the irregular curve of the pattern, as shown by 
gLg. This irregular curve is the only one needed, and is used in 
laying out the patterns for the other sections of the elbow. 










PATTERN PROBLEMS 


39 





































































40 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


The patterns for the sections 2,3 and 4 are usually laid out in 
the following manner: As the patterns are all of equal length, 
from point 1 on each end of the stretchout line HR, draw the 
vertical lines lm. Next, take the wide side of section 2, as 
shown by GF in the elevation, and mark this dimension on the 
lines lm at both ends of the pattern, as shown by gb. 

Then take the throat width of piece 3 and place it as shown 
by ba. The length of long side or top of piece U is placed on the 
same lines, as shown by mo. The rivet holes are spaced, as 
shown by the small circles on the lines lm. 

This completes the drawing, as the patterns are now laid out 
directly on the metal in the following manner: Place the draw¬ 
ing upon the metal, and, by means of a prick punch, transfer 
pattern L to the metal by pricking along the irregular curve. 
The centers of rivet holes and the width of sections are also 
pricked thru the drawing in the usual manner. Pattern L is 
now cut from the metal, after which the metal pattern is re¬ 
versed and the curved edge placed on the points bb. Using a 
scratch awl, the irregular curve is scribed upon the metal, 
which completes the pattern for piece 2, shown at T. 

The patterns for pieces numbered 3 and U are completed by 
placing the curved edge of the metal pattern L on the points aa; 
then scribe the irregular curve on the metal, and draw a line 
from m to m, completing the patterns P and K. This method of 
grouping the patterns places the seams opposite each other and 
allows the patterns to be cut without waste of material. 

Problem 17. Five-Piece 60° Elbow 

Figure 53 shows the elevation of a five-piece elbow, which, 


when completed, should have an angle of 60 °, the inner curve or 
throat being described with an 8-inch radius. 

First, draw the required angle ABC; next, on the line AC, 
measure off a distance of 8 inches from A to G. With AG as ra¬ 
dius and A as center, describe the arc GF. 

Make GC equal 5 inches, and with AC as radius, describe the 
arc CB, which is divided into four equal spaces, one less in num¬ 
ber than the pieces in the elbow. These spaces are shown by 
Be, ee and eC. Bisect these spaces, as shown at m, m, m, m, and 
lines drawn thru these points from the apex will give the miter 
line for each section of the elbow. The end pieces 1 and 5 may 
be made any length, but the length of the heel and throat of the 
middle sections should be taken from the elevation, as shown by 
aa and bb in section 3, and cannot be changed when once the arc 
GF has been described on the drawing. Draw the half section 
K, complete the elevation, and develop the pattern for piece 1 . 

Problem 18. Three-Piece 90° Elbow 

Figure 54 shows the elevation of a three-piece 90 ° elbow for 
which the pattern of piece 1 is to be obtained in accordance with 
principles already explained in Problem 16. 

Problem 19. Profiles of Elbows 

Figure 55 shows the profiles or sections of a square, rectangu¬ 
lar, and oblong elbow, shown at A, B and C. Develop the pat¬ 
tern for a 2-, 3- and 4-piece elbow having a 3-inch throat radius 
and an angle of 90°. The patterns for the end pieces only are 
required. 


SHEET-METAL PROBLEMS 


41 








































































CHAPTER V 

Practical Cornice and Gutter Problems 




The patterns for miters between sheet-metal moldings are de¬ 
veloped by the parallel-line method described in the previous 
chapter, and in order to illustrate the application of the prin¬ 
ciples as applied to moldings, a number of practical problems are 
presented. 

Problem 20. Roman Moldings 

In classical architecture, Green: and Roman moldings are em¬ 
ployed. The outlines of Greek moldings which, in nearly every 
instance, are found to follow the curves of the conic sections, 
generally the parabola or the hyperbola. 

Roman moldings are nearly always formed from the arcs of 
one or more circles, and are chiefly used in sheet-metal cornice 
work. Figure 56 shows a series of Roman moldings, and demon¬ 
strates a method by which each may be geometrically drawn. 

The torus molding, shown at A, is half round and here shown 


between two fillets. It is a semi-circle described from the center 
C, the bisection of the line ab. 

The cavetto, or cove, shown at B, is a concave molding whose 
profile is a quarter circle. The center c is found by extending 
the lines ab and dc until they intersect. 

The ovolo, or quarter round, shown at C, is a convex molding 
with a quarter-circle profile; the center b is obtained in a similar 
manner as the previous molding. 

The cyma recta, known as the ogee, shown at D, is made up of 
two quarter reverse circles tangent at m. The centers g and e 
are found by bisecting the lines ab and cd, as shown. 

The cyma reversa, shown at E, is the reverse of the cyma recta, 
which is concave above and convex below; while the cyma re¬ 
versa is convex above and concave below. The method of con¬ 
struction is similar except the centers g and h are at the top and 
bottom, as shown. 


42 












PRACTICAL CORNICE AND GUTTER PROBLEMS 


43 







Fig. 56. Roman Moldings 









































44 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


The scotia, shown at F, is drawn as follows: Having given 
the points a and b, draw the line ab and bisect ab at c. From c, 
with a radius equal to ca, describe the semi-circle agb. From b, 
draw the line be at an angle of 30 ° to the base line, cutting the 
arc at e. 

From b, erect 
the perpendicular 
bn, and with e as 
center and eb as 
radius, describe an 
arc, cutting the 
perpendicular at n. 

Draw ne and 
drop a perpendic¬ 
ular from a, inter¬ 
secting ne at m. 

With m as center 
and ma as radius, 

draw the arc ahe. _ e 

Problem 21. Square Return Miter. 

With n as center « 

and radius ne, describe the arc eob, completing the molding. 



Problem 21. Square Return Miter 

Figure 57 shows the short method of obtaining the pattern for 
a square return miter. 

This method can only be used when the miter is one of 90°; 
that is, a square miter. Let C represent the profile of a molding 
which is drawn to a scale of 4 inches to 1 foot. Divide the 
quarter round into a number of equal spaces. Number these 
points—also the corners of the molding—as shown by the figures 
1 to 15. The stretchout of the molding may be conveniently 
placed either above or below the drawing of the profile. In this 


case, it was drawn below, as shown by the vertical line AB, from 
which horizontal measuring lines are drawn and intersected by 
vertical lines drawn from similarly numbered points on the pro¬ 
file C. A line traced thru the points thus located will complete 
the pattern, shown atG. The edge lines along which bends are 
to be made are designated by circular indicators in the usual 
way. 

Outside and Inside Miters 

Figure 58 shows the sketch of a roof plan to illustrate the dif¬ 
ference between an outside miter and an inside miter. Miters 
for the outer and inner angles of a roof are called outside and in¬ 
side miters, and are placed as shown in the sketch. 

Pattern G, shown by ABce in Figure 57, is the pattern for an 
outside miter, while the opposite cut, shown by cghe is the pat¬ 
tern for an inside miter. Both patterns are produced by a 
single miter cut, and this is also true when developing patterns 
for miters at any angle. 

Problem 22. Butt Miter 

Figure 59 shows the elevation of a horizontal molding that 
butts against a mansard or other pitched roof, and illustrates 
the principle applicable to butt miters, whether the molding 
butts against a plain or curved surface in the elevation. Let E 
represent the side elevation of a cornice, which is drawn to the 
scale of 4 inches to 1 foot. Draw the profile AB and divide it 
into equal spaces, as shown. From these points on the profile, 
draw horizontal lines parallel to the lines of the molding until 
they intersect the roof line CG. 

At right angles to the lines of the molding, draw the stretch¬ 
out line ab, either above or below the profile. Draw the usual 
measuring lines which are intersected by vertical lines drawn 
from the various intersections on the roof line CG. A line drawn 




PRACTICAL CORNICE AND GUTTER PROBLEMS 


45 



Fig. 57. Square Return Miter 


Fig. 58. Outside and Inside Miters 



































































































46 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


thru these points will be the pattern for the butt miter. Com¬ 
plete the pattern by developing a square return miter cut on the 
end AB. 

Problem 23. An Oblique Return Miter 



Problem 25. Eave-Trough Gutter. 


Figure 60 shows the method of obtaining the pattern for an 
octagon return miter, and is applicable for miters at any angle. 
Patterns for return miters at other than a right angle must be 
developed from a plan of the molding. In this figure, A is the 
profile of a molding which is drawn to a scale 4 inches to 1 foot. 
To draw a plan view of the molding, extend the wall line of pro¬ 
file A, and by the aid of a 45 ° triangle, draw the octagonal angle 
of 135°, as shown by emn, which will represent the wall line in 
plan B. Bisect the angle emn and draw the miter line Gm. 
Divide the curve in the profile into any convenient number of 
spaces and set off the entire stretchout of the profile on the line 
ab, drawn at right angles to the lines of the plan. From all 
points on the profile A, draw vertical lines intersecting the 
miter line Gm in the plan. Measuring lines are drawn from the 
points on the stretchout line ab, which are intersected by hori¬ 
zontal lines drawn from similarly numbered points on the miter 


line Gm. A line traced thru these intersections will complete the 
pattern, as shown at C. The opposite cut of the outside miter 
shown at E, is used should an inside miter be required. 

Problem 24. Butt Miter Against a Surface Oblique in Plan 

Figure 61 shows a horizontal molding at an angle that butts 
against a flat surface, as in the case of a bay window cornice that 
miters against a vertical wall. In this figure, which is drawn to 
a scale 4 inches to 1 foot, cemg represents the face of the wall 
against which the molding A is placed, and GB represents the 
vertical wall against which the butt miter is made. 

Bisect the angles to find the miter lines. At right angles to 
lines of the molding draw the stretchout line ab, and develop 
patterns for moldings H and K. Compare the view here shown 



Problem 26. Molded Face Gutter. 


with that given in Figure 60, and it will be seen that the process 
of development differs but little from that already explained. 

Problem 25. Eave-Trough Miter 

Figure 62 shows the section and pattern for a half-round eave- 








PRACTICAL CORNICE AND GUTTER PROBLEMS 


47 




































































































48 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


trough miter having an angle of 90°, which is drawn to a scale 
of 6 inches to 1 foot. 

The gutter is in the form of a half circle having a bead along 
one of its edges, as shown by the smaller circle. 

Draw the end view of the gutter in the position shown at A. 

Space the outline of 
the end view, as in 
former problems, 
and locate the points 
shown from 0 to H. 

Draw the stretch¬ 
out line ab, and de¬ 
velop the pattern 
for a square return 
miter in the same 
manner as described 
in Problem 21. Pat- 

Problem 27. Ogee Gutter. ternB is the P attern 

for an outside miter, 

while the opposite cut shown at C is the pattern for an 
inside miter. 

Problem 26. Molded Face Gutter 

In Figure 63 is shown a drawing which scales 4 inches to the 
foot, giving the section of a molded face gutter, for which a 
square return miter pattern is to be developed. 

Problem 27. Ogee Gutter 

Figure 64 shows the end view or section of an ogee gutter, 
which is drawn to a 4-inch scale. Develop the pattern for a re¬ 
turn miter having an angle of 135 ° in the plan view. The method 
of development is the same as that shown in Figure 60. 


Problem 28. Molded Roof Gutter 

Figure 65 shows the section of a roof gutter which is drawn to 
a scale of 4 inches to 1 foot. 

Let 1,2, 3, 4, 5 represent the profile of the face, and the line 
5, 6 the back of the gutter. A band iron brace is shown at ce, 
which is bolted to the gutter at a, and secured to the roof by 
nailing at b. Develop the pattern for a return miter having an 
angle of 90°. 

Problem 29. Octagonal Gutter 

In Figure 66 is shown a drawing 4 inches to 1 foot, giving the 
profile of an octagonal gutter. 

Draw the section and plan view and develop the pattern for a 
return miter having an angle of 120°. 


Problem 30. Quarter-Circle Gutter 

The section of a quarter-circle gutter is shown in Figure 67. 



Problem 28. Molded Roof Gutter. 


Obtain the miter line in the plan view and develop the pattern 
for an outside return miter at an angle of 150 °. The profile in 
the drawing is shown one-third full size. 







PRACTICAL CORNICE AND GUTTER PROBLEMS 


49 

































































































































50 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Problem 31. Square-Face Miter (Short Method) 

The method of developing the patterns for a square-face miter 
is shown in Figure 68, and the drawings are made to a scale of 
4 inches to the foot. This process is employed when developing 
the patterns for miters in gable moldings, picture frames, win¬ 
dow and door frames, 
and panel moldings. 

The method of devel¬ 
opment is similar to 
tha't described for the 
square return miter 
(Figure 57). The only 
difference is in the po¬ 
sition of the stretch¬ 
out line ab. In this 
case, the stretchout 
line is placed in a hor¬ 
izontal position at the 
left of profile A, with 
the pattern shown at 
C, while the stretch¬ 
out for the square 
return miter (Figure 57) is placed in a vertical position below 
the profile. 

When developing patterns for square miters, the stretchout 
line must be placed in its proper position, or, instead of having a 
face miter, as shown in Figure 68, the draftsman will have a re¬ 
turn miter, as shown in Figure 57. 

Problem 32. Octagon-Face Miter 

Figure 69 shows the development of an octagon-face miter. 
The patterns for face miters at other than a right angle are de¬ 


veloped by the long method, and the miter line is found in the 
elevation. In this problem the elevation is shown at the right 
of profile A. First, draw the profile A, then draw the required 
angle BGE, which is bisected to obtain the miter line GF. From 
the various points on the profile A, draw horizontal lines inter¬ 
secting the miter line GF, as shown. Draw the stretchout line 
ab at right angles to the horizontal lines of the molding. 

Upon this line place the stretchout of the profile A, and de¬ 
velop the pattern in the usual manner, as shown by amrib. 

The method given in this problem is applicable for miters at 
any angle, and the miter line will be found in the elevation. In 
the case of a return miter, the miter line will be found in a plan 
Anew, as shown in Figure 60. 

Problem 33. Miter Between a Gable and Horizontal 
Molding 

Figure 70 shows another form of face miter in the style of a 
gable molding, shown at AB, which miters with a horizontal 
molding C that butts against an inclined surface, shown by the 
line GF. Draw the elevation, which is shown one-third size 
in the drawing. The miter lines RD and H K are found by 
bisecting the angles in the usual manner. Place the profile E in 
position and develop patterns for the inclined molding B and the 
horizontal molding C. The principles used in developing this 
problem are similar to those given in Problem 32. 

Problem 34. Panel Miter 

Figure 71 shows the elevation of a rectangular panel which is 
drawn to a 3-inch scale. 

Draw a section of the panel mold A, as indicated by the 
shaded portion of the elevation. Divide the cove into a number 
of equal spaces and number each point on the profile. 



Problem 31. Square-Face Miter. 







PRACTICAL CORNICE AND GUTTER PROBLEMS 


51 













































































































52 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


From these points, draw lines parallel to BC, intersecting the 
miter line Cn. From the points on the miter line, draw lines 
parallel to CG, intersecting the miter line Ge. At right angles 
to GC in the elevation, draw the stretchout line ab, and develop 



Problem 34. Panel Miter. 


the pattern for the end of the panel, shown at H. This is the 
only pattern necessary for the construction of the panel, as the 
miter cut on the end pattern is also used for the long sides of 
the panel. 

Problem 35. Roof Finial 

Figure 72 shows the elevation and pattern for a square roof 
finial, which is drawn to a scale of 3 inches to the foot. 
Draw the center line AB and construct the elevation shown at 
C. Divide the profile into a number of equal spaces, and num¬ 
ber the points in the usual manner. 

The finial being square in form, the miter on the corner is sim¬ 
ply a square return miter and is developed by the short method 


shown in Figure 57. Place the stretchout of the profile C upon 
the center line AB, which is extended below the elevation. Draw 
the usual measuring lines, which are intersected by vertical lines 
drawn from points on the profile in the elevation. Now, meas¬ 
uring from the center line, transfer these points to the opposite 
side of the pattern by means of the dividers. 

A line traced thru the points thus obtained will complete the 
pattern for one side of the finial, shown at G. 

Problem 36. 

In Figure 73 is shown the 
half elevation, half sectional 
view, and plan view of a 
square ventilator, which is 
drawn to a scale of 3 inches 
to the foot. Let G represent 
the hood of the ventilator 
and F the flange which is 
joined to the square base, 
shown at G. The band iron 
brace used in connecting 
the hood and base is shown 
at R. The half sectional 
view shows the profile of 
the different sections, and 
in actual shop practice, is all 
that is required for the de¬ 
velopment of the patterns. 

Draw the full size eleva¬ 
tion, omit the plan view, 
and develop the patterns for the different sections by the short 
method shown in Figure 72. 


Square Ventilator 



Problem 35. Roof Finial. 






PRACTICAL CORNICE AND GUTTER PROBLEMS 


53 
























































































































54 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Problem 37. Octagonal Roof Finial 

In Figure 74, A is the elevation of an octagonal finial, the plan 
of which is shown at B. It is drawn one-fourth size. The prin¬ 
ciples explained in the previous problems are also applicable to 
regular poly¬ 
gons having 
any number of 
sides. Each 
angle of the oc¬ 
tagonal figure 
in the plan is 
bisected, and 
the bisectors 
produced until 
they meet in 
the center of 
the octagon. 

The elevation 
shows one of 
the sections, or 
sides, in profile, 
and the plan is 
placed to cor¬ 
respond with 
the elevation. 

Divide the pro¬ 
file into equal 

spaces, and from these points extend vertical lines intersecting 
the miter lines ab and ac in plan. From the center of plan at 
point a, and at right angles to be, draw the stretchout line 
mn thru the points, in which draw the usual measuring lines. 
From the points upon the miter lines ab and ac, draw horizon¬ 


tal lines intersecting corresponding measuring lines. A line 
traced thru these points of intersection will describe the pat¬ 
tern for one of the sides of the finial, shown at C. The pattern 
for one section abc is all that is required. When developing 
patterns for any article, the bases of which are regular poly¬ 
gons, the stretchout line must always be drawn at right angles 
to one of the sides in plan as be, and not at right angles to the 
miter line ab. In actual shop practice, the outlines of the 
elevation are all that are required for the development of the 
pattern. Should it be necessary to draw a finished elevation, 
showing the miter lines in the central portion of the view in Fig¬ 
ure 74, the miter lines 1-e and 1-g are found in the following 
manner: Draw lines parallel to eh in plan from points on the 
miter line ac to corresponding positions on the line ah and ak. 
From these points, vertical lines are drawn, intersecting hori¬ 
zontal lines drawn from similarly numbered points on the profile 
in the elevation. The foreshortened miter lines 1 -g and 1 -e are 
then drawn thru intersections of the vertical and horizontal 
lines. 

Problem 38. Conductor Head 

Figure 75 shows the front and side elevation of an ornamental 
conductor head, which is drawn one-fourth full size. The miters 
on the outer corners are merely square return miters. Draw the 
front and side elevation and develop the pattern for the front of 
the conductor head by the method described in Problem 35. 
The center line divides. the pattern into two equal parts. 
One of these parts will be the pattern for both side pieces, for, in 
this case, the side of the conductor head is equal to one-half the 
width of the front. The front elevation is the pattern for the 
flat back of the head, which is turned in at the top, and the al¬ 
lowance is shown by the dotted lines above the elevation. 








PRACTICAL CORNICE AND GUTTER PROBLEMS 


55 













































































































56 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Problem 39. Pediment Molding Mitering on a Wash 

Figure 76 shows the half front elevation and side elevation of a 
pediment mitering upon a horizontal molding, the roof of which 
is inclined or has what is known as a wash. The drawing is 
made 4 inches to the foot. This wash, or sloping roof, allows 
rain and snow to drip off, and 
should always be placed when 
the pediment molding has 
great projection. 

Draw the center line AB, 
and on this line place the width 
of each member of the hori¬ 
zontal molding, as shown by 
FEDCB; also the distance to 
the top of the pediment, shown 
by G. At right angles to AB 
draw the lineFH equal to one- 
half the width of the pediment. 

From H erect a vertical line 
intersecting the line C, or bot¬ 
tom of the wash at point 3. 

Draw a line from 3 to G on 
the center line, which gives 
the pitch of the pediment. 

Draw the line 2-10 at right angles to the pediment mold, and 
construct the profile as shown at K. Divide the cove into 
equal spaces, shown from U to 7, thru which lines are drawn 
parallel to the lines of the molding, intersecting the center line 
AB from 1 to 10, as shown. 

Draw the side elevation and construct the profile of the hori¬ 
zontal molding, shown by abcdefg. The line of the wash is 
shown by cb. From c and b draw vertical lines, as shown, and 


intersect them by a horizontal line drawn from point G, which 
completes the side elevation of the top of the pediment;. 

Before the pattern for the inclined molding can be described, 
it will be necessary to obtain the elevation of the miter between 
the inclined mold and the wash. To obtain this miter, a duplicate 

of the profile at K is placed in 
the side elevation in the posi¬ 
tion shown at P. From the 
points on profile P, draw ver¬ 
tical lines intersecting the wash 
cb, as shown. Now, from the 
points of intersection on cb, 
horizontal lines are drawn to 
the front elevation, which in¬ 
tersect lines drawn from cor¬ 
responding points on the pro¬ 
file K. A line traced thru the 
points of intersection, as shown 
from a to 3, will complete the 
elevation of the miter between 
the wash and the inclined 
molding. 

The pattern for the inclined 
molding K is shown at R, and 
is developed in the usual manner. 

Draw the stretchout line mn at right angles to the pitch of the 
pediment mold, upon which place the girth of profile K, as 
shown by the numbers 1 to 10. Thru these points draw the 
usual measuring lines, which intersect by lines drawn at right 
angles to G-3 from similarly numbered points on the miter lines, 
at the top and bottom of the oblique mold. 

A line traced thru these points will give the required pattern, 



Problem 39. Pediment Molding Mitering on a Wash. 







PRACTICAL CORNICE AND GUTTER PROBLEMS 


57 














































































58 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


to which is added the triangular piece shown by 10-B x -l 0 X in the 
elevation. Using B-10 and B-10” as radii, with 10 and 10 x in 
the pattern as centers, describe arcs intersecting at B x . Draw 
lines from 10 to B x , to l(f, completing the pattern. 

The half pattern of the horizontal mold with the miter cut in 
the wash is shown at N. To develop the pattern, place the 
stretchout of the horizon¬ 
tal molding on the center 
line AB, as shown. From 
these points draw horizon¬ 
tal lines, which intersect 
by vertical lines drawn 
from similarly numbered 
points on the miter line in 
the elevation. 

A line traced thru the 
points thus obtained will 
be the pattern for the hor¬ 
izontal molding. 

The pattern for the two 
end pieces of the horizon¬ 
tal molding are obtained 
by taking a duplicate of the profile L, shown by chsf, to which 
laps are added on all sides. 

Problem 40. Gable Molding with Raked Profile 

The gable ends of buildings are frequently finished with mold¬ 
ings similar to the upper members of the main cornice. A mold¬ 
ing so used is called a raked molding. When an inclined molding 
is to miter with a horizontal return, a change of profile in one or 
the other of the moldings is required, to insure a correct miter of 
the moldings. In Figure 77 is shown the front and side eleva¬ 


tion of a gable finish or pediment, mitering with a horizontal 
return molding at the bottom, at a right angle in plan. In this 
case, the profile of the return at the bottom is the original, or 
normal, shown at A, while that of the inclined molding is raked, 
or changed to correspond, as shown at B. First, draw the nor¬ 
mal profile A and divide it into equal spaces in the usual manner. 

From these points draw 
the lines of the horizontal 
molding and the inclined 
molding, as shown. 

To obtain the raked pro¬ 
file B, draw the horizontal 
line be below the normal 
profile A. From the points 
on profile A, draw vertical 
lines intersecting the line 
be, and number the divi¬ 
sions, as shown, from 3 to 
9. Next, take the various 
divisions on be, and place 
them on the line b'c', which 
is drawn parallel with the 
lines of the oblique molding. From the various intersections and 
at right angles to b'c', draw lines intersecting similarly numbered 
lines drawn from corresponding points in profile A. A line 
traced thru the points of intersection thus found will give the 
profile of the raked molding. The lower part of profile B that 
miters on the wash is next drawn, and the divisions located 
upon the line b'c', as shown. 

The top of the raked molding extends back to the wall, and 
the projection is shown by the line 2-3. The stretchout line mn 
is next drawn at right angles to the rake molding, upon which 



Problem 40. Gable Molding, with Raked Profile. 











































60 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


place the stretchout of profile B, as shown from 1 to 13. Each 
** space of the curved portion of the profile must be taken separate¬ 
ly, for the divisions are of unequal length. 

Next, draw the usual measuring lines which are intersected by 
lines drawn from points on the normal profile A, and the miter 
line EF, thus com¬ 
pleting the pat¬ 
tern, shown at G. 

The pattern for 
the return at the 
bottom of the 
raked molding is a 
square return mi¬ 
ter, which miters 
upon both the 
raked and horizon¬ 
tal moldings, and 
is developed from 
the profile A in the 
usual manner. The 
elevation of the 
miter between the 
raked molding and 
the Wash —also the Problem 41. Broken Pediment. 

pattern for the horizontal molding—are obtained in the same 
manner as described in the previous problem. The operations 
are shown in Figure 76. The profile of the horizontal molding 
is shown by abcdefgh in the side elevation. Observe that the 
inclined wash shown by the line cb is not extended to the wall 
line HK. 

Problem 41. Broken Pediment 

A pediment is an ornamental form of gable finish for door and 


window openings, and is frequently placed over the top of the 
main cornice of a building. 

The half front and side elevation of a form of this kind, known 
as the broken pediment, is shown.in Figure 78, which is drawn 
to a scale of 4 inches to the foot. The drawing shows a section of 
the horizontal molding and one arm of the inclined molding, which 
is cut off some distance below the center, and does not extend to 
the upper miter as in Figure 77. The open space at the top of a 
broken pediment is usually filled in with some ornamental form. 

This problem shows the method used to rake the return mold¬ 
ings when the normal profile is used for the inclined molding, 
and the process is exactly opposite that shown in Figure 77. 

First, draw the lineG# to the required angle of the inclined 
molding. The normal profile is then drawn in the position 
shown at A, and is divided into equal spaces in the usual man¬ 
ner. From these points on the profile and at right angles to 
G H, draw lines intersecting the line ran, which is drawn parallel 
to GH. ■ Next, take the various divisions from 3 to 10 on the 
line mn and place them on the horizontal line m'n', as shown. 

From the divisions on m'n ' draw vertical lines intersecting 
similarly numbered oblique lines drawn from corresponding 
points in profile A. A line traced thru the points of intersection 
will give the raked profile B. 

To find the raked profile C at the upper end of the inclined 
molding, establish the point 3' on the line G H and take the va¬ 
rious divisions from 3 to 13 on mn and place them on the hori¬ 
zontal line m z n z , having the point 3 placed directly over 3'. 
From the divisions on the line m z n z , draw vertical lines, inter¬ 
secting similar lines in the inclined molding. The raked profile 
C for the upper return molding is then traced thru the points of 
intersection. 

The pattern for the inclined molding is shown at F, and the 





PRACTICAL CORNICE AND GUTTER PROBLEMS 


61 





cr a 

































62 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


method of development is similar to that described in Problem40. 

The pattern for the upper and lower returns are simply square 
return miters, and further description is unnecessary. 

The profile of the horizontal molding is shown at E in the side 
elevation, the slope of the wash is represented by the line cb, and 
extends back to the wall line R K. 

The method of finding the miter between the inclined and 
horizontal moldings—also the cut 
on the wash of the horizontal mold¬ 
ing—is fully shown in Figure 76. In 
actual shop practice, it is unneces¬ 
sary to draw a side elevation of the 
inclined molding shown at P to ob¬ 
tain the points 12-13 on the wash cb. 

The required points can be found in 
the following manner: At any con¬ 
venient distance above the line cb, 
representing the wash, and at right 
angles to the wall line R K, draw the 
line m s n s . Take the divisions 10-11, 

12-13 and Y-2 on the line mn, and 
place them on the line m s n s , as shown, being careful to place the 
point Y-2 on the wall line R K. Vertical lines drawn from these 
points to the line cb will give the required points on the wash 
without drawing the side elevation of the inclined molding. 
Problem 42. Miter Between Moldings of Different Profiles 

It is often necessary in cornice work to bring moldings of dif¬ 
ferent profiles together in a miter. To develop patterns for a 
square return miter between moldings of dissimilar profiles re¬ 
quires two distinct operations, and the miter cut upon each 
piece would be the same as it would appear when a butt miter 
was made between the two moldings. Figure 79 shows the pro¬ 


files and patterns for a square return miter between moldings of 
two different profiles, which are drawn one-third full size. The 
two arms of the miter being different in profile, it is necessary to 
draw an elevation of the two moldings in the position shown at 
A and B. The profiles having been drawn in the position 
shown, the patterns are developed in the same manner as the 
butt miter, shown in Figure 59. Divide the profile A into any 
convenient number of parts, from 
which draw horizontal lines inter¬ 
secting profile B, as shown. To 
develop the pattern for molding A, 
place the stretchout of the molding 
upon the line ab, which is drawn 
at right angles to the lines of the 
molding. 

Draw the usual measuring lines 
which are intersected by vertical 
lines drawn from similarly num¬ 
bered points on profiled. 

A line traced thru these points 
will give the miter cut of the pat¬ 
tern for molding A, to fit against the profiles. 

The pattern for an outside miter is shown at C, and an inside 
miter at E. To obtain the pattern for molding S, proceed in the 
same manner, reversing the order of the profiles. Develop the 
stretchout of molding S, on the line mn, being careful to take 
each division separately, as they are of unequaled width. Draw 
the horizontal measuring lines which are intersected by ver¬ 
tical lines drawn from similarly numbered points on profile A. 

A line traced thru these points will give the pattern for mold¬ 
ing S, which miters with profile A. An outside miter pattern is 
shown atF, while the opposite cut is an inside miter, shown atG. 



Problem 42. Miter Between Moldings of Different Profiles. 




PRACTICAL CORNICE AND GUTTER PROBLEMS 


63 




Fig. 79. Mitering Moldings of Different Profiles 























































































































































































































































































































PART THREE 

Radial Developments 


chapter VI 

Regular Tapering Forms 


This subject embraces a large variety of forms of frequent 
occurrence in sheet-metal work, and the forms or shapes con¬ 
sidered within this part include only such forms that have for 
their base the circle or any of the regular polygons, as the 
square, hexagon, octagon, etc.; also figures, tho of unequal 
sides, that can be described within a circle, in which lines drawn 
from the corners terminate in an apex over the center of the 
base. , 

When developing patterns for tapering forms, the following 
rules will enable the student to understand the principle by 
which these developments may be accomplished. 

1) A drawing must first be made consisting of an elevation 
showing the true height of the apex, and the true length of the 
radius with which to describe the stretchout of the pattern. 

. 2) A plan view must be drawn from which the length of 


the stretchout can be obtained, as shown by EF in Figure 80. 

3) The stretchout arc must be described with a radius equal 
to the length of the true edge of the solid, as shown by AH 
Figure 80. 

4) Points are located on the stretchout corresponding to the 
position of the points on the outline of the plan or sectional view. 

5) Measuring lines and edge lines of the pattern are always 
radii of the stretchout arc. 

6) Should an irregular or straight line be drawn through any 
cone, as illustrated by the line G-9 in Figure 82, and which the 
radial lines in elevation will intersect, then, from these points of 
intersection on G-9, lines must be drawn at right angles to the 
axis A H intersecting the side of the cone AB, which will give 
the true lengths from apex A, and are carried to similarly num¬ 
bered radial lines in the pattern R, as shown from 1 to 1. 


64 












REGULAR TAPERING FORMS 


65 




































66 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Problem 43. Pattern for Cone and Frustum 

In Figure 80 is shown the method of developing the pattern 
for a right cone. This method contains the principles applicable 
to all pyramids which have for their base the circle or any of the 
regular polygons that can be inscribed within a circle. 

First draw the center line AB, upon which place the height of 
the cone as AC; thru C draw the horizontal line GH equal to 
the width of the base, and draw lines from G and H to A. Di¬ 
rectly below the elevation, 
describe a circle to represent a 
plan view of the base as shown 
by EF, which is divided into 
equal spaces as shown from 
1 to 9. 

The radial lines shown in the 
elevation, plan and pattern are 
not necessary, but are shown 
to make clear their relation to 
each other. To obtain the pat¬ 
tern for the cone, use the apex 
A as center and with a radius 
equal to the true length of the Problem 45^ Irregular Frustum 
slant height of the cone as 

shown by A H, describe the stretchout arc HM. On any con¬ 
venient point on the stretchout locate point 1 and draw a line 
from 1 to A. Then set off on the arc HM of the pattern, com¬ 
mencing at 1 , twice the number of spaces that are contained in 
the half circumference of the plan, as shown by 1-9-1. From 
the end point thus located draw a line to the apex A, and 
then add an allowance for seaming. If a frustum of a cone is 
desired, as shown by GDRH, then using AR as radius, draw 
the arc RP, making R-P-l-1 the desired pattern. 


Problem 44. Pattern for a Square Pyramid 

This development is shown in Figure 81. Draw the plan and 
elevation according to the dimensions given in the drawing. 
When the plan view is placed in the position as shown, the line 
CB in the elevation represents the true length of one of the cor¬ 
ners of the pyramid; the stretchout arc may, therefore, be de¬ 
scribed as in the case of the cone in the preceding problem. With 
C as center and CB as radius, describe the arc BG. After setting 
the dividers to the width of one side of the base shown in the 
plan at 1-2, begin at 1 and step off on the stretchout arc BG 
spaces equal in number to the sides of the pyramid. Thus, 
points are located at 1,2, 3, U and 1. Connect these points by 
straight lines as shown, and draw lines from each point to the 
apex A, completing the drawing. The outline C-l-2-S-ly-l is 
the development of the pyramid. 

Problem 45. Pattern for an Irregular Frustum of a 
Cone 

Figure 82 shows the method of developing the pattern for the 
frustum of a right cone cut by the plane represented by the line 
G-9 drawn oblique to the axis of the cone, which also intersects 
the radial lines from 1 to 9. 

First draw the elevation of the cone ACB, and directly below 
it the plan view D. Divide one-half of the plan into equal 
spaces as shown by the figures 1 to 9. Next draw the line G-9 
at an angle of 45° with the base line CB, making the point 
G one inch from C. From the various points in the plan erect 
lines intersecting the base of the cone from 1 to 9, and from 
these intersections draw lines to the apex A, intersecting the line 
G-9 as shown. 

Using A as center and with AB as radius, describe the stretch¬ 
out arc BF, upon which step off twice the number of spaces 





REGULAR TAPERING FORMS 









































68 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


shown in the plan D, as shown by similar numbers 1 -9-7. From 
these points draw radial lines to the apex A. From the various 
intersections on the line G-9, and at right angles to the axis line 
AH, draw lines as shown intersecting the side of the cone AB 
from 1 to 9. Then, using A as center with radii equal to the va¬ 
rious divisions as A-l, A-2, A-3, etc., draw arcs intersecting 
similarly numbered radial lines in the pattern R. The irregular 
curve traced thru points thus obtained completes the desired 
development. 

Problem 46. Pattern for an Irregular Frustum of a 
Hexagonal Pyramid 

This development is shown 
in Figure 83, the plan and 
elevation being first drawn 
according to the dimensions 
given in the figure. The 
cutting plane is shown by 
the oblique line mn and is 
drawn at an angle of 45° with 
the base line. 

The true length of the edge 
lines in the pyramid are not 
shown in either the plan or 
elevation, and it is, therefore, 
necessary to draw a line that 
will represent the true edge 
in the elevation. This is found 
as follows: From a as center of the plan view, with the 
radius a-5, describe the arc 5-c, intersecting the line ab at c. 
From c draw a vertical line to the base line BC of the front 
elevation extended at E. AE is then the true length of one of 


the comers of the pyramid, and is also the true length of the 
radius of the stretchout arc. 

The stretchout arc EG is next described from A as center with 
a radius AE. The widths of the sides of the base are then laid 
off at 1-2-3-1*-, etc.; connect these points by straight lines, as 
shown, and draw lines from each point to the apex A. 

From the intersections on the oblique line mn and at right 
angles to the axis AF, draw lines to the true edge line of the pyra¬ 
mid at ejg. With A as center and radii Ae, Af and Ag, describe 
arcs intersecting similarly numbered radial lines in the pattern, 
shown at H. 

Complete the development by drawing lines connecting these 
points in the manner shown. 

Problem 47. Conical Gutter Outlet 

In many cases of eave-trough construction, where it is desired 
to so connect the conductor pipe that the opening in the gutter 
will be larger than the diameter of the pipe, a tapering connec¬ 
tion pipe is used, as shown in Figure 84. 

An examination of the drawing shows that the flaring outlet 
consists of a frustum of a cone; its upper base in the drawing is 
defined by the straight lineF H, and the lower base by a section 
of the gutter. First, draw the center line AB and construct a 
section of the gutter, shown by the arc CGD. Make the distance 
from G to 0 equal to the height of the outlet, and thru 0, at right 
angles to the center line AB, draw the lineF H equal in length to 
the diameter of the conductor pipe. 

Draw the line mn, intersecting the center line at R. Then, 
using R as center, describe the half plan of cone, as shown by 
men. From m and n, draw lines thruF and H, intersecting the 
center line at A, completing the elevation of the entire cone. 

Develop the patterns for the conical outlet mFHn, using 



Problem 46. Irregular Frustum of 
a Hexagonal Pyramid. 




REGULAR TAPERING FORMS 


69 











































70 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


principles similar to those used in Problem 45, in the develop¬ 
ment of an irregular frustum of a cone. 

Problem 48. Tapering Collar for Roof Having a Double 

Pitch 

The principles used in developing the pattern for the inter¬ 
sected cone in Figure 82 are applicable, no matter at what 
angle or point the bases of 
the cone are intersected. 

Develop the patterns 
for a tapering collar, 
shown by C HmGD in Fig¬ 
ure 85, the base of which 
is to fit against a roof of 
two inclinations, as indi¬ 
cated by DGE; also when 
it has a single pitch, as 
shown by FE. The dot¬ 
ted line Dm represents 
the base, and A the apex 
of the cone. The half 
plan is shown at K, be¬ 
low the elevation. 

Problem 49. Rectangular Pyramid 

Figure 86 shows the development of a pattern for a rectangu¬ 
lar pyramid. This principle is applicable to various-shaped or¬ 
naments in cornice work; also all pyramids that have for their 
base any of the regular polygons that can be inscribed within a 
circle, the lines drawn from the corners of which would termi¬ 
nate in an apex directly over the center of its base. 

Patterns for work of this kind are usually laid out directly 


upon the metal by a method in which no elevation is required, as 
the true length of the radius for describing the stretchout arc is 
found in the plan. The length, width and height of pyramid 
being known, first draw a plan to the required size, as shown by 
1-2-3-U, and then the diagonal lines 1-3 and 2-k, which repre¬ 
sent the miter lines or hips, intersecting in the center at a. Bisect 
the line 1~U and locate the point e; then draw a line from e to a, 

showing the position of the 
seam. 

Before describing the 
stretchout arc for the pat¬ 
tern, the true length of 
one of the hip lines in plan 
must be found, and that 
dimension used as the ra¬ 
dius for describing the 
stretchout arc. 

To find the true length 
of the hip line a-2 in plan, 
erect the perpendicular 
a-b equal to the vertical 
height AM in the eleva¬ 
tion. Draw a line from 2 
to b; then 2-b is the true 
length of the line 2-a in the plan, and is the rsdius with which 
to develop the pattern. With any point with F as center, 
describe the circle g-d. 

Starting at any convenient point on the arc, as point 2, step 
off the length of the end and sides of the pyramid, shown by the 
divisions 1-2-3 -4, and draw lines to the apex F. From 1 and U 
as centers with a radius equal to one-half the width of the base, 
as shown by 1-e in plan, describe the short arcs, as shown. The 



Problem 47. Conical Gutter Outlet. 





REGULAR TAPERING FORMS 


71 















































72 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


true length of the seam line is shown by the dotted lineFw. Then 
with Fn as radius and F as center, describe arcs at bh. Connect 
all points by straight lines, completing the pattern. 

Problem 50. Octagonal Pyramid 

From Figure 87, draw the plan view and develop the pattern 
for an octagonal pyramid. Apply 
the method used in Problem 49. 

Problem 51. Oblong Pitched 
Cover 

Figure 88 shows the method of 
developing the pattern for an ob¬ 
long raised cover with semi-circu¬ 
lar ends. First draw the plan and 
elevation, which will show that the 
shape consists of the two halves of 
the envelope of a right cone, con¬ 
nected by a straight piece in the 
center. To develop the pattern, 
proceed as follows: Draw the line 
AB equal in length to CG of the 
plan. From A and B as centers, 
with radius equal to L H of the elevation, describe the stretch¬ 
out arcs, as shown by am and bn. Upon these arcs, from a 
and b, step off twice the number of spaces which are con¬ 
tained in one-half the semi-circular end, as shown by the divi¬ 
sions 1 to 5 in plan, thus obtaining the points m and n. From 
m draw the line ma, and from n draw nB. From A and B, at 
right angles to the center line AB, draw the line Ae and Bh 
equal in length to Am of the pattern. 

Connect e and h, completing the development. 


Problem 52. Flaring Pan 

In Figure 89 is shown the elevation and half pattern for a 
flaring pan made in two pieces, the form of which is seen to be 
the frustum of a right cone. An inspection of the drawing will 
show that AB, the top of the pan, is the base of an inverted cone, 
its apex C being at the intersection of the lines AG and B H 
forming the sides of the pan, and 
that G 7/ is the top of the frustum 
or the base of another cone which 
remains after taking the frustum 
from the original cone. In devel¬ 
oping the pattern, first draw the 
center line FC, upon which place 
the height of the pan FR; thru 
these points draw lines at right an¬ 
gles to the center line. On either 
side of the center line from the 
points FR, place the half diameter 
FB of the top and HR of the bot¬ 
tom. Draw lines connecting BH 
and FA and extend them until they 
meet the center line in the point 
C. With R as center and RH as 
radius, describe the quarter circle HD, and divide it into a 
number of equal parts, as shown by the figures 1 to 7. This 
quarter-circle represents a one-quarter plan of the bottom of the 
pan. The pattern is developed as follows. With C as center 
and the radii equal to CH and CB, draw the arcs OE and K N, 
as shown. From the center C, draw a line across these arcs 
near one end, as KO, and starting from the point K, step off on 
the arc K N twice the number of spaces contained in the quarter 
plan, as shown by the figures 1-7-1 on the arc. Thru the last 



Problem 52. Flaring Pan. 




REGULAR TAPERING FORMS 


73 


































































74 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


division, draw a line across the arcs to the center C, as shown by 
E N. Add laps for wiring and seaming, as shown by the dotted 
lines. A one-half elevation and a quarter plan of the top or bot¬ 
tom is all that is required to find the stretchout and radius for 
describing the pattern for the frustum of a cone. 

Problem 53. Pattern for a Funnel 

Figure 90. Applying the method given in Figure 88, develop 
the patterns for the body 
and spout of the funnel, 
shown in the drawing. The 
vertical height is 3-1/2 
inches. The diameter of 
the top is 5 inches, and the 
lower opening in the body 
measures 1 inch in diame¬ 
ter. The spout is 2 inches 
long and has a 1/2-inch 
outlet, the seam being 
lapped and soldered. It 
will be seen from Figure 90 
that all the operations 
required may be per¬ 
formed if only one-half of 
the elevation be drawn, 
as shown on one side of the vertical center line AB. 

A one-quarter plan of the top and bottom is shown in the half 
elevation by the quarter-circles which are described from m and 
n as centers. As a matter of convenience, the pattern for the 
body, shown atG, is laid off to one side of the elevation, and the 
stretchout arc is not here described from the apex of the cone as 
was conveniently done in the previous problem. The pattern 
for the spout is shown at C. 


Problem 54. Round Ventilator Head 

In Figure 91 is shown the half elevation and half section of a 
round ventilator head. The conical hood is shown at A, and the 
lower flange at F. The supports C and G are made from band 
iron riveted to the hood A and the round pipe £, as shown in the 
drawing. The lower flange F has an inclination of 45°, and the 
pitch of the upper hood A is at an angle of 30°. Draw the full 
elevation and develop the patterns for the upper hood A and the 
flange F. 

Problem 55. Fruit-Jar Filler 

Figure 92. Develop the patterns for a fruit-jar filler. The 
flaring body is shown at A, the spout at B, and the handle at G. 
The pattern for the handle is shown atG, to which an allowance 
has been added for a 3/16-inch hemmed edge on each side. 

Problem 56. Elliptical Flaring Pan 

The plan, elevation and. half pattern for an oval flaring pan are 
shown in Figure 93. As may be understood from the drawing, 
the bases are elliptical, while the sides flare uniformly. The 
short and long diameters of the lower base or top of the article 
are 9 x 12 inches, respectively. The sides flare 1-1/2 inches all 
around; that is, the upper base or bottom is an ellipse, whose 
long and short diameters are 6 and 9 inches, respectively. The 
vertical height is 3-1/2 inches. Draw the plan view according 
to the rule given in Figure 26 (Practical Geometry) and locate 
the centers a, b, e, m, as shown. Next, draw the elevation 
BFGH, setting off the vertical height 3-1/2 inches. The next 
step is to obtain the radii with which to strike the arcs of the 
pattern. 

From b and m in the plan, draw the vertical lines bA and mE, 
and extend the side HF of the elevation until it intersects the 
perpendiculars from b and m in the points A and E. Then A H 



Problem 54. Round Ventilator Head. 












REGULAR TAPERING FORMS 


75 










































































































76 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


is the radius with which to strike the pattern for that part, 
shown by 13-7-a in the plan, and AF the radius for that part, 
shown by 7-1-m in plan. With AH as radius and R as center, 
describe the arc cn in pattern K. Starting at any point as c. 


set off on the arc cn, the stretchout of the arc 13-7 in plan. 
Draw lines from 13-7 to R. 

Then with EH in elevation as radius and n in pattern K as 
center, describe an arc cutting the line nR in P. 

With P as center and Pn as radius, describe the arc 7-1, upon 
which set.off the stretchout of 7-1 in plan. Draw a line from 1 
to P. Next, with radii equal to AF and EF in the elevation and 
with centers R and P, describe the arcs dh and hg, thus com¬ 
pleting the pattern with seams at 1 and 13. Allowances for 
seaming and wiring must be added to the pattern. 

Problem 57. Grocer’s Scoop 

In Figure 94 is shown the elevation and patterns for a hand 
scoop, the body being in the form of an intersected cylinder, and 
the handle and brace are the frustums of two right cones. 

This problem as presented will require the development of 


patterns by both the parallel line and radial methods, and the 
student’s attention is directed to the manner in which any 
irregular section of the cylinder may be produced. 

First draw the side elevation and half section to the dimen¬ 
sions given in the drawing. Divide the half section G into a 
number of equal spaces, shown from 1 to 7, and from these 
points draw horizontal lines intersecting the outline of the side 
elevation. To obtain 
the pattern for the 
body F, draw the 
stretchout line AB in 
the position shown and 
proceed to develop it 
in the usual way, the 
spaces shown by 1 -7-1 
being set off in the 
usual manner. 

The handle H is the 
frustum of a cone 
shown by cmae, which 
is soldered to the flat 
back of the scoop. The 
conical boss or brace 
is shown by hngb. 

The patterns for the 
brace and handle are 
shown at R and C, and the method of development has been 
fully described in previous problems. 

Problem 58. Tapering Square Pipe Intersecting a Vertical 
Square Pipe 

Figure 95 shows the manner of developing the pattern for a 
tapering square pipe intersecting a vertical square pipe placed 



Problem 56. Elliptical Flaring Pan. 











REGULAR TAPERING FORMS 


77 






































78 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


diagonally in plan. In this problem, first draw the plan view of 
the vertical pipe in the position shown by GDTE; directly above 
it, the elevation A. Next, in its proper position in elevation, 
draw the tapering pipe egch, extending the lines of the pipe 
equally until they intersect in the apex H and the base line ab. 

Below the base line ab, draw the section F, as shown. The top 
and bottom corners of the tapering pipe intersect the corner of 
the vertical pipe in the plan at G, shown in the elevation by c 
and h, while the side corners 
intersect the flat sides of 
the vertical pipe at 2~U in 
the plan and elevation, and 
this point must be found as 
follows: From the apex H 
in elevation, drop a vertical 
line intersecting the center 
line of plan at R. Next, 
take the distance m-2 and 
m-U in section F, and from 
B in center of plan, place 
this dimension upon the line 
DE, as shown at f and d. 

Draw the lines Rd and Rf, 
cutting the sides of the ver¬ 
tical pipe at 2 or 4- From these points, draw a vertical line 
intersecting the center line HO in elevation at 2'-p. Connect 
ch and 2'~U' by straight lines, which will give the miter or line 
of intersection between the two pipes. Before the pattern for 
the tapering pipe can be developed, the true length of the cor¬ 
ners must be found as follows: From the intersections 1' and 
2'-lt' at right angles to the center line HO, draw lines until 
they intersect the line eh, the lower corner of the tapering pipe. 


To obtain the pattern for the pipe, use H as center, and with 
Ha as radius, describe the arc ok. After setting the dividers to 
the width of one side of the base, shown in section F by 1-2, 
begin at 1 and step off on the stretchout arc ak, spaces equal in 
number to the sides of the pipe, and from these points draw 
radial lines to the apex H. With H as center and radii equal to 
H-l s , H-3 1 and H-2*-lf, draw arcs intersecting similar radial 
lines in the pattern P, as shown. Connect the various points, 
and the desired pattern is obtained. 

Problem 59. A Cone Intersected by a Cylinder Placed 
Vertically 

Figure 96 illustrates the principles for developing a vertical 
cylinder intersecting with a cone. The construction of this 
problem must be followed very carefully, as several of the opera¬ 
tions are necessarily made over one another on the drawing, and 
the student must be careful to distinguish each process. 

First draw the elevation of the cone ABC, in accordance with 
the dimensions given on the drawing. The plan is then drawn, 
as shown, and thru the center a draw the diameter FR. Estab¬ 
lish the location of the center of the cylinder at g, and with g as 
center, describe a circle, which will represent the outline of the 
vertical pipe, of the diameter indicated on the drawing. Divide 
the upper half of the circle into equal spaces, as shown by the 
figures 1 -1^-7. Thru each of these points on the circle, from a as 
center, draw the half circles, as shown, intersecting the center 
line FR from 7 to 1. 

These half circles represent the plan views of horizontal planes 
which are projected to the elevation. Then from points 7 to 1 
on the center line FR, draw vertical lines intersecting the side of 
the cone AC from 1' to 7'. 

From each of the points of intersection with the side of the 






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80 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


cone, draw horizontal lines which are intersected by vertical 
lines drawn from similarly numbered points on the outline of the 
cylinder in plan. A line traced thru these points, as shown, will 
be the miter line. 

From the points 1 and 7 on the miter line, draw the vertical 
lines 1-G and 7-H, 
which connect from G 
to H, completing the 
elevation of the cylin¬ 
der. 

The pattern for the 
cylinder, shown at D, 
is obtained in the man¬ 
ner usual with all 
parallel forms; the 
stretchout is taken 
from the spaces upon 
the outline of the cyl¬ 
inder in plan, which 
are transferred to the 
line mn, as shown. The method for obtaining the pattern for 
the cone is similar in process to Figure 80. 

A section of the pattern, and the method for obtaining the 
opening to fit the cylinder, is shown at E. 

With A as center and radii equal to A-7', A-6', A-5', A-tf, 
A-3', A-2' and A-l ', describe arcs, as shown. Draw the center 
line AP. Then, measuring from the center line FR in plan, take 
the various distances along the arcs from the center line to the 
points on the outline of the small circle, representing the cylin¬ 
der in plan, and place them on similar arcs in the pattern E, 
neasuring on both sides of the line AP. A line traced thru these 
points, as shown, will give the pattern for the opening in the cone. 


Problem 60. A Cone Intersected by a Vertical, 
Square Pipe 

Using the plan shown at A, Figure 97, and the same size eleva¬ 
tion as in Figure 96, develop the pattern for a cone intersected 
by a vertical square pipe placed diagonally, as shown in the plan 
at m. 

The principles used for developing the patterns in Figure 96 
are also applicable to various problems, no matter what the pro¬ 
file or form of the pipe may be. 

Problem 61. A Cone Intersected by a Cylinder at Right 
Angles to Its Axis 

Figure 98. The principles in this problem do not differ from 
those given in Figure 96. Draw the elevation of the cone ABC 
and the plan view F. Next, draw the elevation of the cylinder 
G-3-3-H; also its section, shown by R, which is divided into 
equal spaces, shown from 1 to 5. From these points, draw hori¬ 
zontal lines intersecting the side of the cone AC. 

Then, from the points of intersection on the line AC, draw 
vertical lines intersecting the center line ab in plan at 2~U, 1-5 
and 2-1+. With F as center and radii equal to F-2-J+, F-l -5 and 
F-2-1+, describe the circles shown. Extend the center line ab in 
plan and draw a duplicate of profile R, as shown by D, changing 
the position of the numbers shown from 1 to 5 to 1. From the 
points on profile D, draw horizontal lines intersecting similarly 
numbered circles, as shown by 1' to 5' in plan. From these in¬ 
tersections, draw vertical lines, which intersect similarly num¬ 
bered horizontal lines in the elevation. A line traced thru these 
points will give the miter line between the cone and cylinder. 

The pattern for the cylinder can be developed from the inter¬ 
sections in the plan or elevation in the usual manner. The pat¬ 
tern for the opening in the cone is shown at K, and is obtained 





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82 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


in the manner explained in Problem 59. Measuring on either 
side of the center line ab, take the distances along the arcs in 
plan and place them on similar arcs in the pattern, measuring 
from the center line AP. A line traced thru the points thus ob¬ 
tained will give the pattern for the required opening in the cone. 


same manner as the points on the profile of the cylinder in 
Figure 98. 

Obtain the miter line nbd in the elevation formed by the junc¬ 
tion of the square pipe and the cone, and develop the patterns 
for the cone and the horizontal square pipe. 


Problem 62. A Cone Intersected by a Square Pipe Placed 
in a Horizontal Position 


Problem 63. The Frustum of a Cone Intersecting a 
Cylinder Obliquely 


Figure 99 shows the plan and elevation of a cone intersected 
by a square pipe at right 
angles to its axis. 

The principles illus¬ 
trated in Figure 98 for 
obtaining the intersec¬ 
tions between a cone and 
cylinder placed horizon¬ 
tally are also applicable 
to problems where the 
pipe is square or elliptical 
in form. Draw the plan 
view of the cone, and thru 
the center m draw the di¬ 
ameter eg. Directly 
above the plan, draw the 
elevation ABC, and in its 
proper position, the ele¬ 
vation of the square pipe 
ahdn; also the profile 
shown by F, which divide into equal spaces to obtain the 
points a and b. 

These points are also located upon the profile H in plan, 
and are projected to the plan and elevation of the cone in the 



Problem 63. Frustum of Cone Inter¬ 
secting a Cylinder Obliquely. 


The principles given in this problem are applicable to cone 
and pipe intersections at any angle, and the cone may be placed 
in the center or to one side of the pipe. This method can also be 
applied to problems where the vertical pipe is square or rectan¬ 
gular in form. 

Figure 100 shows the plan, elevation and patterns for the frus¬ 
tum of a cone intersecting a cylinder of greater diameter than 
itself at an angle of 45 °. The drawing is made to a scale of 6 
inches to 1 foot. 

Draw the elevation of the cylinder AFCB, and the plan view, 
shown at G. Locate the point R and draw the elevation of the 
cone Rba, extending the sides of the cone until the base line 7-1 
is obtained. From the center e on the base line, draw the half 
view of the base, which divide into equal spaces, as shown 
from 1 to 7. From the various points on the half view of the 
base and parallel to the center line of the cone, erect lines inter¬ 
secting the base of the cone, and from these intersections draw 
radial lines to the apex R. 

Thru the center of the cylinder in plan, parallel to the base 
line EC in elevation, draw the line R'H, which is intersected at R' 
by a vertical line projected from R in the elevation. 

With m as center, draw a full view of the base of the cone, and 
number the divisions from 1 to 7, as shown in D. 

From these points, draw horizontal lines, which intersect by 




REGULAR TAPERING FORMS 


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Fig. 100. Frustum of Cone Intersecting Cylinder 






































































84 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


vertical lines drawn from similarly numbered intersections on 
the base line 7-e-l in the elevation. 

From these intersections in plan, shown by l'-2'-3', etc., 
draw lines to the apex R', intersecting the plan of the cylinder 



Problem 64. Frustum of Cone Intersecting 
a Cone Obliquely. 


from a to /. From the intersections a to /, draw vertical lines 
which intersect corresponding radial lines in the elevation. A 
line traced thru these points, as shown, gives the line of inter¬ 
section or miter line between the cone and cylinder. 

The points of intersection on the miter line are now projected 
at right angles to the center line of the cone to the true edge line 
Ra, as are also the points at the intersection of the radial lines 


with the horizontal line nh which represents the upper base of 
the frustum. With R as center and R-7 as radius, describe the 
stretchout arc, and develop the pattern for the frustum of the 
cone in the regular way. 

The pattern for the opening in the cylinder, shown at K, is 
obtained by the parallel-line method; the stretchout line EC ex¬ 
tended is drawn in the position shown, and the spaces a, b, c, d, e 
and /, as found in the plan, are spaced off on .that line. Thru 
the points thus located, vertical lines are now drawn, and inter¬ 
sected by horizontal lines from similarly numbered points in the 
line of intersection in the elevation. 

Problem 64. The Frustum of a Cone Intersecting a Cone 
of Unequal Diameter Obliquely 

Figure 101 shows the method of obtaining the patterns for in¬ 
tersecting cones, and the drawing is made to a scale of 4 inches to 
1 foot. Draw the elevation of the larger cone as ABC, and in its 
proper position below same, describe the circle E, which repre¬ 
sents the plan view of the base. Next, thru point b on the line 
AB, draw the center line of the smaller cone at an angle of 45° 
with the base of the larger cone, and locate the points F and m. 

Thru m at right angles to the center line Fm, draw the line 
5-m-l. With m as center, describe the semi-circle 1-3-5 1 
which represents a half view of the base of the smaller cone. 
Divide the outline into a convenient number of equal parts, 
shown from 1 to 5, and from these points draw lines to the base 
line 5^nt-l. Radial lines are now drawn from these intersec¬ 
tions to the apex F. Divide the quarter-circle 6-8 in plan E into 
equal spaces, as shown by the figures 6-7-8, and draw radial 
lines to the center n, as shown. In practical work, it will be 
found necessary to use a larger number of spaces in order that 
the line of intersection may be more accurately traced. From 




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86 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


the intersections 6-7-8 in plan, draw vertical lines intersecting 
the base line BC, from which radial lines are drawn to the apex A, 
intersecting the radial lines of the smaller cone F-5-1, as shown. 

A series of sections of both cones are next drawn in the plan. 
The sections of the smaller cone will, in each case, be a triangle, 
while the sectional curves of the larger cone will be elliptical, 
parabolic or hyperbolic curves, as the case may be. From the 
point of intersection with the side of each triangle and its cor¬ 
responding sectional curve, vertical lines are drawn which inter¬ 
sect similarly numbered radial lines of the smaller cone, and the 
line of intersection between the two cones is traced thru these 
points. 

The sectional triangles of the smaller cone are obtained in the 
following manner: From the points 5, W, 3', 2’ and 1, on the 
base line of the smaller cone, draw vertical lines to the plan, and 
set off distances from the center line GC similar to the distances 
from 5-m-l in the half view of the base; that is, make S'-3 
equal h-3 s , etc. From these points, draw lines to the apex G, 
completing the triangles. 

The next step is to obtain the sectional curves of the larger 
cone on the planes F-V, F-S' and F-2'. The extreme upper and 
lower points 1 and 5 are the points of intersection of the central 
section, and are projected directly to the plan from the eleva¬ 
tion. Where the plane F-Y crosses the base line BC at e and 
the radial lines drawn from 6' and 7', shown by dbg, lines are 
projected into the plan until they intersect similar radial lines 6 
and 7; also the pla-n of the base at e'. The irregular curve a'e’ 


traced thru these points of intersection is a section of the larger 
cone produced by the intersection of the cutting plane ea in ele¬ 
vation. 

In similar manner obtain the sectional curves thru the planes 
bu and gd, as shown in plan by b'u' and g'd', the distance from n 
to d', and n to u' being obtained from u to v and d to r, respec¬ 
tively, in the elevation. 

From the points of intersection in plan where the sectional 
triangles of the smaller cone intersect the sectional curves of the 
larger cone, vertical lines are drawn which intersect similar 
radial lines in the elevation at 1,2,3,U and 5, as shown. 

From these intersections at right angles to the center line 
Fm, draw lines intersecting the side of the cone F-5. 

With F as center and F-5' as radius, describe the stretchout 
arc and develop the pattern for the frustum of the cone in the 
usual manner, shown at R. 

A section of the pattern and the method for obtaining the 
opening in the larger cone is shown at H. 

With A as center and AC as radius, describe the arc CP. 
Next, draw any line as A-l-5, on either side of which place the 
various divisions 1+, 2, 3, which are taken from the arc 6-8 in 
plan and placed upon the arc CP, as shown. 

From these points, draw radial lines to the apex A, which in¬ 
tersect by arcs drawn from A as center with radii equal to the 
divisions on the side of the cone, as shown on the line AC. 

A line traced thru the intersections shown from 1 to 5, will 
give the pattern for the opening in the larger cone. 



CHAPTER VII 

Triangulation 


There are numerous irregular forms arising in sheet-metal 
work, the patterns for which cannot be developed by the regular 
methods employed in the two previous chapters. These irreg¬ 
ular shapes are so formed that, altho straight lines can be drawn 
upon them, the lines would not run parallel to one another, nor 
would they all incline to a common center. In parallel-line de¬ 
velopments, the distance between any two lines running with 
the form is the same at both ends of the article, while in radial 
developments, all lines running with the form tend toward a 
common center or apex, so that the distances between such lines 
at one end of the article (provided it does not reach to the apex) 
govern those at the other end. 

Hence, in the development of the pattern for any irregular 
form, it becomes necessary to drop all previous methods, and 
simply proceed to divide the drawing representing the surface 
of the article into triangles. 

Then from the drawing, the true lengths of the various sides 
must be found, and the triangles constructed therefrom. 


The construction of a triangle whose three sides are given is 
not a difficult problem, and it becomes a simple problem in 
geometry to construct the triangle. Having found the true 
lengths of the sides of such triangles, reproduce them in regular 
order in the pattern, and, hence, the term triangulation is most 
fittingly applied to this method of development of surfaces. 

Problem 65. Transition Piece, Square to Square 

While irregular forms are largely curved surfaces, the method 
of triangulation is best illustrated by its application to a form 
having plane surfaces, as shown in Figure 102. Both bases are 
square, and, in this case, parallel, but diagonally-arranged in 
their relation to each other, as may be seen from the drawing. 
Draw the elevation and plan view, in which a, b, c, d represents 
the square base, and 1,2, 3, U the plan view of the square top, 
each side of which shows its true length. Now, connect the top 
and base by drawing lines from the corners in plan, as shown. 
These lines represent the bases of the triangles, which must be 


87 






88 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


constructed so as to find the true lengths of these lines. This is 
accomplished by constructing, in each case, a right-angled tri¬ 
angle, whose base is equal to the length of any fore-shortened 
line in the plan, and its altitude to the vertical height of the 
same line shown in the elevation. The hypotenuse of such a 
triangle will then be equal to the true length of the line. In this 
case, the lines b-1 , C-2, d-3, 
a-4, etc., are all represented 
by lines of the same length. 

The vertical height AB is the 
same in the case of each line. 

A triangle constructed by 
the above method will be 
sufficient to indicate the true 
length of these lines, and is 
constructed as follows: Draw 
any horizontal line as mn, 
and from m erect the perpen¬ 
dicular mh equal to the alti¬ 
tude AB in the elevation. 

As all of the lines connect¬ 
ing the corners in plan are equal, only one triangle is necessary. 
Therefore, take the distance from b to 1 in plan and place it, 
as shown, from m to n, and draw the line hn, which represents 
the true length of the lines b-1, c-2, d-3, etc., in plan. 

The pattern is to be laid out in one piece, with a seam thru 1-e 
in plan. 

Take this distance 1-e and place it, as shown, from m to g, and 
draw a line from h to g, which will be the true length of the seam 
line 1 -e. The true lengths of all the lines in plan having been 
found, the triangles may now be placed in position in the pat¬ 
tern, care being observed that the adjacent triangles are 


completed in the same order as they are shown on the solid. 

Draw any horizontal line as cd in the pattern, equal to cd in 
the plan. Now, with radius equal to hn in the triangles and cd 
in the pattern as centers, describe arcs intersecting each other at 
3 . Draw lines from c to 3 and 3 to d. Then c-d-3 is the correct 
development of the surface c-3-d in plan. The adjacent tri¬ 
angles c-2-3 and d-3-k may next be constructed. With 3-2 in 
the plan as radius, and 3 in the pattern as center, describe the 
arcs 2 and 4; these arcs are then intersected by arcs described 
from c and d as centers, with a radius equal to hn in the diagram 
of triangles, thus developing the triangles c-3-2 and d-3-U, 
which correspond to similarly numbered surfaces in the plan. 
Now, with radius equal to cb and da in plan, and c and d in pat¬ 
tern as centers, describe the arcs b and a, which intersect by arcs 
described from 2 and U as centers and hn in the triangles as 
radius. Draw lines from 2 to b to c and 4 to a to d in the pattern, 
which is the pattern for the sides d-U-a and c-2-b in plan. In 
like manner develop the surface of the figure shown by k-a-l 
and 2-b-l in the plan. Then, with radius equal to ae in plan, 
and a and b in the pattern as centers, describe the arcs mm, 
which intersect by arcs described from 1 and 1 as centers and hg 
in the diagram of triangles as radius. Draw the lines b to m to 1 
and a to m to 1, completing the pattern. 

Problem 66. Register Box, Rectangular to Round 

Figure 103 shows the manner of developing the pattern for a 
register box whose top is rectangular and base is round, and may 
also be described as a transition piece; that is, a form used to 
connect outlines unlike in shape. A fitting of this kind is fre¬ 
quently used in the sheet-metal trades, particularly in the con¬ 
struction of bases for chimney tops, ventilator heads and fan 
connections in blow-pipe work. 



Problem 65. Transition Piece, 
Square to Square. 




TRIANGULATION 


89 












































90 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


The elevation, plan and half pattern shown in Figure 103 are 
drawn to a scale of 3 inches to the foot. First, draw the 
rectangle abed, which represents the plan of the top, and circle 
e-5-l-g, which shows the plan of the base of the article. As the 
circle is in the center of the rectangle, making the four quarters 
symmetrical, it is necessary only to divide 
the one-quarter circle into a number of equal 
parts, as shown by the figures 1, 2, 3, b, 5, 
from which draw lines to the corner a. These 
lines will form the bases of a series of tri¬ 
angles whose altitude is equal to the vertical 
height of the article, and whose hypotenuses 
will be the real distances from a in the base 
to the points assumed in the curve in the top. 

Next, draw the elevation ABCD, and add 
the straight flange and collar to the top and 
base, as shown. The next step is to find the 
true lengths of the lines 1-a, 2-a, 3-a, etc., in 
plan. To construct a diagram of triangles, 
first draw the line G H, and from G lay off 
the distances, shown by the lines in plan, 
thus making G-l equal to a-1, G-2 equal to 
a-2, etc. At right angles to HG, draw GF, 
in height equal to the straight height of the article, as shown 
in the elevation, and connect the points on the line G H and F. 
Also set off the distance m-5 from G, and draw the line F-n, 
which will give the true length of the seam line m-5 in plan. 

To develop the half pattern, first draw any line, as a'b', equal 
in length to ab in plan. Now, with F-l of the diagram of tri¬ 
angles as radius and a' and b' as centers, describe arcs intersect¬ 
ing each other, thus establishing point 1 of pattern. Next, 
with the points a' and b' as common centers and radii equal 


to the true lengths of the lines a-1, a-2, a-3, etc., of the plan, 
as shown in the diagram of triangles, describe arcs of indefi¬ 
nite length. Set the dividers to the length of one of the 
spaces on the quarter circle in plan, and, commencing at 
the point 1 in pattern, step off a number of spaces on each 
side to correspond to those shown on the 
quarter-circle from I to 5 in plan. Thru 
the points thus obtained, trace a line, as 
shown from 5 to 6. With a' and b' of pattern 
as centers, and am of plan as radius, describe 
short arcs, which intersect other arcs, de¬ 
scribed from 5 and 5 as centers, and Fn of 
diagram of triangles as radius, thus estab¬ 
lishing the points m' of the pattern. 

Now, connect the various points by draw¬ 
ing lines from 5 to m' to a' and 5 to m' to b’, 
completing the half pattern for the tapering 
body of the register box, shown by ABCD in 
the elevation. The vertical flange R and 
laps for seaming are added, as shown in the 
pattern. The lower collar, shown at P, is 
simply a straight piece of pipe for which a 
pattern is not required, as it can be laid 
out directly upon the metal. 

Problem 67. Register Box, Rectangular to Round, 
Vertical on Two Sides 

Figure 104 illustrates a condition occasionally met by the 
sheet-metal worker engaged in erecting stacks, furnace hoods, 
register boxes, and various other fittings used in heating and 
ventilating work. The principles in this problem do not differ 
from those given in the preceding problem, and are applicable, 



Problem 66. Register Box, Rectangular 
to Round. 






TRIANGULATION 


91 






































































52 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


whether the round opening of the article is placed exactly in the 
center of the base or at one side or corner. 

Figure 104 shows a drawing (3 inches to the foot) of a 
register box whose top is rectangular, and whose base is round 
and placed in one corner of the rectangle, as shown. First, 
draw the rectangle abed, 
which represents a plan view 
of the top, and the circle 1- 
5-9-13, which shows the plan 
of the round base of the arti¬ 
cle. As the circle is not in 
the center of the rectangle as 
in the previous problem, the 
four quarters are not sym¬ 
metrical, and it is necessary 
to divide the entire circle 
into a number of equal parts, 
as shown, from which draw 
lines to the corners a, b, c and 
d. Next, draw the elevation 
ABCD and add the straight 
flange G and the round collar 
F, as shown. The true lengths 
of the lines in each quarter 
of the plan are found by constructing the diagram of triangles, 
shown at a', b', e' and d', by the method described in the 
previous problem. Having found the true lengths of all lines 
in plan, develop the full pattern for register box, placing the 
seam at m-1 , as shown in plan. 

Problem 68. Irregular Flaring Pipe Connection, Vertical 
on One Side 

The style of pipe-fitting shown in Figure 105 is frequently 


used in blow-pipe-fitter's work, especially where the lower side 
of a round duct is required to be perfectly straight, to avoid any 
obstruction to the material that is being forced thru the pipe by 
means of a fan. 

Draw the plan and elevation in accordance with the dimen¬ 
sions given. In the plan view, after drawing the horizontal cen¬ 
ter line, shown by 13-1 , divide the outline of each of the half sec¬ 
tions into the same number of equal parts. Number these 
points, as shown, from 1 to 13, and draw lines connecting the 
successive points. The true lengths of all lines in the plan must 
now be determined by means of a diagram of triangles, and the 
development of the pattern for the irregular flaring section, 
shown by ABCD in the elevation, will differ in no material re¬ 
spect from the development explained in Problem 66. The sec¬ 
tions F and G are simply straight pieces of pipe for which a pat¬ 
tern is not required. 

Problem 69. Transition Piece, Round to Oblong 

Figure 106 shows the plan view and side elevation of a pipe 
fitting or transition piece used in connecting an oblong and 
round pipe. The sheet-metal worker engaged in heating and 
ventilating work has frequent use for a fitting of this kind, and it 
is generally known as a center boot, which is used to connect an 
oblong riser or wall pipe with a round horizontal pipe. The 
round end of the boot is connected to the pipe by means of a 
three- or four-pieced elbow which gives a gradual turn, making 
direct connection with the furnace hood or air chamber in hot¬ 
air heating work. 

In constructing this drawing, the plan is to be drawn first. 
Draw the center line AB and construct the outlines of the upper 
and lower bases. Draw the vertical center line CG, which 
divides the plan into symmetrical quarters; all the work that is 



Problem 68. Irregular Flaring 
Pipe Connection. 





TRIANGULATION 


93 

























































94 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


necessary for the development of the pattern may, therefore, be 
accomplished in one of these divisions. Next, draw the side 
elevation of the transition piece, as shown by F HDR, and place 

the straight collars E 
and P in position, as 
shown. Divide the 
quarter-circle represent¬ 
ing the upper base into 
a number of equal parts, 
shown by 2, 4, 6, 8, and 
an equal number of 
spaces are set off on the 
quarter-circle represent¬ 
ing the lower base, 
shown by 1, 3, 5, 7. 
Draw lines connecting 
the successive points in 
the plan, shown by 1-2, 
2-3, 3-4, etc. The true 
lengths of these lines 
are next determined by 
means of a diagram of 
triangles constructed by 
the method previously shown for such cases, and a one-half 
pattern is developed, placing the seams in the position shown by 
A and B in the plan. The student being already familiar 
with the method of procedure, no further instruction is 
necessary. 

Problem 70. Twisted Rectangular Pipe 

An elevation and plan view of a twisted rectangular pipe 
is shown in Figure 107. The profiles of the upper and lower 


ends are alike, and they are placed diagonally one above the 
other, as shown in the plan. In order that the plan be more 
readily understood, the plan of the bottom of the pipe is 
represented by the numbers 1, 2, 3 and 4, and the top end 
by the letters a, b, c and d. Connect the various corners 
and find the true lengths of the lines by the usual method 
that has been explained in similar triangulation problems 
already given. Develop the pattern in one piece, placing the 

seam in the position shown by the 
line eb in the plan view. 

Problem 71. Transition Piece, 
Rectangular to Triangular 

Figure 108 shows a transition 
piece, the bottom of which is tri¬ 
angular in form, the top being a 
rectangle. Let 1, 2, 3, 4 and 5 
represent the plan of the top, and 
a, b and c the plan of the bottom 
of the transition piece. Draw the 
plan and elevation and develop 
the pattern in one piece, placing 

Problem 70. Twisted Rec- the seam at the comer, shown by 
tangular Pipe. 0-1 in the plan. 

Problem 72. Irregular Fitting, Forming a Transition 
from Round to Oblong. Upper Base Inclined 45° 

In Figure 109, AB-1-13 represent the side elevation of an 
irregular fitting, whose top opening, shown by the line 1-13, 
does not run parallel with the lower base. The first step is to 
draw a plan view showing the outline of the oblong base with 
semi-circular ends. Draw the center line 2-m-l 6, thus dividing 




Problem 70. Twisted Rec¬ 
tangular Pipe. 






TRIANGULATION 


95 


the plan into symmetrical halves. From m draw a vertical line, 
and locate the point e three inches above the base line AB in the 
elevation. Thru the point e, draw the line 1 -13, the angle of in¬ 
clination being 45°. With e as center, describe the half-circle 
representing the half profile of the upper base. Next, divide 
the half-profile c into a number of equal parts, and from these 
divisions at right angles to the upper base, draw lines intersect¬ 
ing the line 1-13, as shown. 

Divide the lower base of the fitting, as shown in the plan, into 
a number of equal parts. Next, take a tracing of the half pro¬ 
file C and place it in the 
position shown at F, 
numbering the divisions, 
as shown, from 1 to 7. 

From the points on the 
half-profile F, draw hori¬ 
zontal lines, which inter¬ 
sect by vertical lines drawn 
from the points 1-3',-5', 
etc., on the upper base in 
the elevation. A line 
traced thru these points 
of intersection will give 
the foreshortened view of 
the upper base in plan. 

Draw lines connecting the 
points on the upper and 
lower bases in plan, as in the 
preceding problems, for 
the purpose of defining the 
triangles. The true lengths of the lines 1-2, 2-3, 3-1+, etc., are 
found in a slightly different manner from that used in the pre¬ 



vious problems, in which the upper and lower bases were parallel 
with each other. In this case, the vertical distances are not the 
same and the true lengths of the lines in plan are found in the 
following manner: Extend the lower base line AB in the eleva¬ 
tion to the right and left, as 
shown on the drawing, and on 
these lines set off the lengths 
of the lines 1-2, 2-3,3-k, etc., 
as they appear in the plan at 
G and H. Draw the vertical 
lines 1-n and 7-h, and upon 
these lines the vertical heights 
are projected from the eleva¬ 
tion, as shown. 

The true lengths of all the 
lines having been determined, 1 
develop the full pattern, plac¬ 
ing the seam on the line 13- 
16, as shown in the plan view. 

The collar, shown by 1-R- 

P-13, is "simply a short piece of round pipe, the pattern being 
laid out directly upon the metal. 



Problem 72. Irregular Fitting, Tran¬ 
sition from Round to Oblong. 


Problem 73. Irregular T-Joint 

In many cases of blow-piping, it is necessary to connect two 
pipes of different diameters at various angles. An ordinary T- 
joint would not suffice, for the reason that it is important to se¬ 
cure an easy flow of air thru the pipes. To accomplish this re¬ 
sult, an irregular flaring connection piece is often used, as shown 
in Figure 110. The upper base is round and is rightly inclined 
at an angle of 30°. The lower base is oblong in form and is a 
portion of a cylinder. It is desired to so construct the fitting 






96 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


that the opening in the large pipe will be somewhat larger than 
the diameter of the smaller pipe. As may be seen from Figure 
110, the method of development does not differ materially from 
that of the preceding problem. First, draw the plan view of the 
oblong base, noting that, in this case, the outline is a foreshort¬ 
ened view of the real 
surface. Next, draw a 
line vertically upwards 
from C, the center of 
the plan, to the point a 
in the elevation. 

Thru a draw the line 
1-13 at an angle of 30° 
with the horizontal; 
with a as center and 
a-1 as radius, describe 
the half circle repre¬ 
senting the one-half full 
view of the upper base, 
as shown. With b on 
the vertical center line 
as center and a radius 
of 3-1/2 inches, describe 
in the elevation the arc geh, representing the view of the 
lower base and a section of the large pipe. Since the view of 
the oblong base in the plan is foreshortened, it is necessary 
to draw a full view, in order to ascertain the true distance 
around the lower base. To produce the full view, shown at A, 
extend the center line 2-16 in the plan toward the right, and on 
this line lay off the stretchout of the lower base, as shown by the 
numbers 2', U', 6', 8', etc., on the arc geh in the elevation. 
From these points on the stretchout line in the full view, draw 


vertical lines, which intersect by horizontal lines drawn from 
similarly-numbered points in the plan. A line traced thru these 
points will give the true outline of the lower base of the fitting; 
measurements are taken from points on this outline for the 
stretchout of the lower edge of the pattern, their true lengths 
thus being shown. The true lengths of these lines are found by 
constructing a diagram of triangles on both sides of the eleva¬ 
tion. Two diagrams are constructed to avoid confusion from 
having a number of lines cross on the drawing. The . true 
lengths of all lines having been found, the pattern may be 
developed by methods precisely like those used in preceding 
problems. 

Problem 74. Roof Collar, Square to Round, 

Base Obliquely Inclined 

Figure 111 illustrates this problem. In this figure, 1-9-a-b is 
the elevation of a roof collar having a round top and square 
base, when viewed on a horizontal line. The collar fits over an 
inclined roof having a pitch of 45 °, shown by the line CG in the 
elevation. First, draw the plan in accordance with the dimen¬ 
sions given on the drawing. Next, draw the elevation and con¬ 
struct the diagram of triangles, as shown at e. The true 
lengths of all the required distances may now be taken from 
their respective places on the drawing and a full pattern con¬ 
structed in exactly the same manner as has been explained in 
similar problems already given. 

Problem 75. Scalene or Oblique Cone 

The development of this problem (shown in Figure 112) illus¬ 
trates a short method of triangulation applicable to a number of 
problems, particularly to those represented by transition pieces, 
the bases of which can be inscribed within a circle. Figure 112 





TRIANGULATION 


97 
















































































98 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


shows the elevation, half plan and pattern for an oblique cone, 
and an inspection of its radial lines will show that they are of 
unequal length. The development cannot be made by the 
method applied to radial solids, altho the process is a com¬ 
bination of that method and 
triangulation. 

Let AHB represent the ele¬ 
vation, and BFH the half plan, 
drawn for convenience so that 
the line HB serves as both the 
base line of the elevation and 
the center line of the plan. 
From the vertex A, draw a 
vertical line intersecting the 
baseline HBatC. Divide the 
half circle BFH into equal 
parts, shown from 1 to 7, and 
draw lines to the apex C. 
These lines will form the bases 
of a series of right-angled tri¬ 
angles of which AC is the altitude, and whose hypotenuses will 
show the lines in their true length. The most convenient 
method of constructing these right-angled triangles is to trans¬ 
fer the distances from C to the various points in the half plan, 
to the base line HB, measuring each time from the point C. 
Then, with C as center and radii equal to C-2, C-3» C-U, etc., 
draw arcs intersecting the base line HB. 

Lines are drawn from each of these points to the apex A, thus 
producing the elements of the oblique cone in their true length, 
the method used being similar to that used in the other triangu¬ 
lation problems. Now, using A as center, with radii equal to 
A-l', A-2', A-3', A-h', etc., draw arcs, as shown. From any 


point upon the arc drawn from point 1 ', as m, draw the line 1-A. 
Now, set the dividers equal to the spaces contained in the half 
circle in the plan view, and starting from the point 1-m, step 
from one arc to another, as shown from 1 to 7, after which com¬ 
plete the opposite half to 1-x. A line traced thru these points, 
as shown, will be the required pattern. To obtain the pattern 
for the frustum of an oblique cone, shown by GR HB in the ele¬ 
vation, use A as center, and with radii equal to the various 
divisions on GR, draw arcs intersecting similar radial lines in the 
pattern. The curve eab traced thru these points of intersection 
completes the development. 

Problem 76. Oblong, Raised Cover With 
Semi-Circular Ends 

The principle given in the preceding problem is applicable to 
the development of the oblong raised cover with semi-circular 
ends, shown in Figure 113. 



Problem 76. Oblong, Raised Cover. 


On examining the drawings, it will be seen that those parts of 
the surface bounded by the curved outline and sloping to the 
point A' in the plan are portions of an oblique cone. First, 
draw the plan view, the semi-circular ends being described from 











Fig. 112. Scalene or Oblique Cone 


TRIANGULATION 


99 



Fig. 113. Oblong, Raised Cover 






































100 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


the centers b and e. Draw a horizontal center line thru the 
plan, and divide one-quarter of the curved outline into a con¬ 
venient number of parts, as at 1, 2, 3, U and 5. From these 
points, draw lines to the apex A' in the plan, and let mA' repre¬ 
sent the seam line of the cover. Next, draw the elevation, the 
height of the cover being shown by the center line AB. From 
A' in plan as center, describe arcs, 
respectively, from 5, 4, 3 and 2, pro¬ 
ducing them until they reach the 
horizontal center line; from these 
intersections, draw vertical lines until 
they reach the base line of the cover 
in the elevation. Lines drawn from 
these points to the apex A will then 
represent the true lengths of similarly 
numbered lines in the plan. To 
obtain the half pattern, shown at G, 
use A as center, and with radii equal to A-l, A-2, A-3, A~k, 
A-5 and A-m', draw arcs, as shown. From point 1 on the 
outer arc, draw the center line 1-A in the pattern. Then, with 
the dividers set equal to the various divisions in the quarter 
plan m-1 , set off spaces on corresponding arcs in the pattern, on 
either side of the line 1-A. Trace a line thru the various inter¬ 
sections, thus completing the half pattern for the raised cover. 

Problem 77. Rectangular Raised Cover With Rounded 
Corners 

^ In Figure 114 is shown an elevation and plan view of a rec¬ 
tangular raised cover with quarter-circle corners which is drawn 
to a scale of 3 inches to the foot. 

The shape of the cover may, perhaps, be more accurately de¬ 
scribed as that of an oblong pyramid with rounded corners. An 


inspection of the drawing will show that the rounded corners are 
portions of a scalene cone, while the four sides are simply flat 
triangular surfaces. 

First draw the plan view; the rounded corners being described 
from the centers a, e, g and c. Next, divide one of the quarter 
circles into a number of equal parts and from these divisions, 
draw lines to the apex A. Draw the 
elevation BFG and develop a half 
pattern by the method described in 
Problem 76. Let the line mAm in 
plan represent the seam line of the 
cover. 

Problem 78. Six-Pointed Star 

Figure 115 shows a plan view and 
elevation of a six-pointed star, which 
is drawn to a scale of 6 inches to the 
foot. The development of this problem does not differ materi¬ 
ally from that in the two preceding problems. First, draw the 
plan view of the star, the height of which is equal to Am in the 
elevation. The line ab in plan is shown in its true length by 
the line AC in elevation. The line eb is also shown in its true 
length in the plan. The true length of the line ae in plan is 
obtained by taking the length of this line as radius, and with a 
as center, describe an arc intersecting the horizontal line ab at g. 
From point g, draw a vertical line intersecting the base line BC 
in the elevation, as shown at n. Draw the line An, which will 
show the true length of the line ae in plan. 

Having found the true lengths of all lines required in the de¬ 
velopment, construct a half pattern, with the seam on the line 
Oe in plan. A section on the line oh in plan is shown by the out¬ 
line okh and is obtained in the following manner: Extend oh 



Problem 77. Rectangular Cover with Rounded Comers. 




TRIANGULATION 


101 























































102 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


until it intersects the base line BC in elevation at t, from which 
point at right angles to AB, draw ts. Place the distance ts from 
d to k in plan. Connect ko and kh, which will give the true pro¬ 
file of the desired section. 

Problem 79. Ash or Garbage Chute Head 

Ash or garbage chute heads are made in many different styles 
dependent on the pleasure of the draftsman. The main point 
to be observed in producing a design for a chute of this kind is 
to see that all unnecessary angles and elbows are omitted, so 
that no obstructions are placed in the way of the contents. 

A plan and elevation of a chute head that answers these re¬ 
quirements is shown in Figure 116. The body of the head is 
constructed from one piece of metal, with seams on AB and 4-C, 
as shown in the drawing. The plan is first to be drawn in ac¬ 
cordance with the dimensions given. First, describe the circle 
to represent the round pipe, and then draw the rectangle that 
represents the plan view of the opening in the body of the chute. 
The horizontal center line is now drawn thru the center of the 
circle and divides the rectangle into symmetrical halves. Next, 
represent the outline of the round pipe in the elevation, and 
draw the vertical center line. From the center point 1' on the 
base line B-b', draw the line 1 '-G at an angle of 45° to the base 
line. Placing the triangle opposite to the position it had when 


drawing the line 1 '-G, move the triangle along the line 1 ’-G and 
locate the point C four inches from G and outside the line U U' of 



Problem 79. Ash or Garbage 
Chute Head. 


outline of the completed 
shown at F. 


the round pipe. Draw the 
lines C-4 and G-V, complet¬ 
ing the elevation of the body. 
Short collars are connected at 
the top and bottom, as shown 
by e-b-k-A in the elevation. 
Divide the outline of the 
round pipe in the plan into a 
number of equal spaces, and 
from these points draw verti¬ 
cal lines to the elevation inter¬ 
secting the upper and lower 
base lines, as shown. 

The true lengths of all lines 
in the plan and elevation may 
now be found by construct¬ 
ing a diagram of triangles by 
the method previously ex¬ 
plained, and the pattern de¬ 
veloped in one piece. An 
rn, which is reduced in size, is 









CHAPTER VIII 
Triangulation 
Simplified Method 


This chapter explains the simplified method of developing 
patterns by triangulation, in which no plan view is required, the 
elevation simplybeingused,on either ends of which are placed 
the semi-profiles, from which the altitudes for obtaining the true 
lengths of the lines in the elevation can be found. 

This method can be used only when both halves of the article 
are alike or symmetrical. The elevation must always be drawn 
at a right angle to a line drawn thru the center of the plan, 
which divides the article into two symmetrical parts. 

Problem 80. Irregular Flaring Roof Collar 

The application of the simplified method of triangulation for 
developing patterns for irregular forms in sheet-metal work is 
shown in Figure 117, which is drawn one-third full size. The 
drawing shows the pattern and elevation for an irregular taper¬ 
ing roof collar that fits over a pitched roof, indicated by the line 
mn. A full view of the upper and lower bases of the collar will 
be exact circles. The form of the collar, altho closely resem¬ 
bling the frustum of a regular cone, is such that its pattern can 
be developed only by triangulation. First, draw the vertical 


center line 1^-1 V in the elevation, and thru the point IV, draw 
the slanting roof line mn at an angle of 30°. 

Next, with a radius equal to one-half the diameter of the lower 
base and with the point IV on the roof line mn as center, de¬ 
scribe the half circle representing a half profile on the lower base, 
as shown at C. The upper base and half profile, shown at B, is 
next drawn; also the lines 1-1 h and 7-8, completing the outlines 
of the elevation shown at A. Divide the half profile B of the 
upper base into a number of equal spaces, as shown from 1 to 7, 
and draw a vertical line from each point at right angles to the 
base line 1-7, intersecting the line 1-7, and numbering each 
point to correspond with each number upon the half profile, as 
shown by 2', S', V, etc. In like manner, divide the half profile 
C of the lower base into the same number of equal spaces, as 
shown from 8 to lk, and from each point at right angles to the 
base line mn, draw lines intersecting mn, and number each point 
of intersection to correspond with the points on the half profile 
C, as shown by 9', 10', IV, etc. Connect the points on the up¬ 
per and lower bases by solid lines, as shown by 2'-13', S'-12', k'- 
IV, etc. Also connect points on the base line mn with points 


103 






104 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


on the. upper base line 1-7 by dotted lines, as shown by IS'-S', 
12'-If', 11'-5', etc., thus dividing the surface of the collar into 
triangles. 

These solid and dotted lines in elevation show the base line of 
sections which will be constructed, the altitudes of which are 
equal to the heights in the 
two half profiles. If this 
elevation with the two half 
profiles were cut from 
cardboard or thin metal, 
and the half profiles B and 
C turned up at right angles 
on the lines 1 -7 and mn, we 
would have a model of one- 
half the article, as shown 
in the illustration at F. 

If wires or threads were 
placed in position, con¬ 
necting the points .on the 
outline of the upper and 
lower profiles, we would at 
once have the true lengths 

of the solid and dotted Problem 80. Irregular Flaring 
r . ,, , ,. .„ Roof Collar, 

lines in the elevation as if 

measured upon the surface of the finished article. The true 
lengths of the solid and dotted lines are shown in the diagrams 
G and H, and are obtained as follows; 

Take .the distances in elevation of the solid lines 13'-2', 12'-3', 
11' -V, 10'-5', 9'-6', and, measuring from a, place them on the 
horizontal line in G, as shown by similar numbers. From these 
points at right angles to the horizontal line a-6', draw the ver¬ 
tical lines 2'-2, 3'-3, Jf'-lf, etc., making each in length equal to 


the distance of points of corresponding numbers in the half pro¬ 
file B of the upper base from the center line 1-7, as measured 
upon the lines at right angles to the line 1-7, thus obtaining the 
points 2, 3, If, etc. Upon the horizontal line a-6', erect a per¬ 
pendicular, as a-11. Upon a-11 set off from a the several dis¬ 
tances of the points in the half profile C from the center line 1 lf-8 
of the lower base, as indicated by the figures 11, 10-12, 9-13. 
Now, connect these points with points on the vertical lines by 
means of solid lines, as shown. Then 11-U, 10-5, 9-6, etc., will 
represent the true lengths of similar solid lines in the elevation. 

The true lengths of the dotted lines are shown in diagram H 
and are obtained in a similar manner from measurements in the 
elevation. The horizontal line llf-7 is a duplicate of a-6, upon 
which set off the lengths of the several dotted lines in the eleva¬ 
tion. From each point draw vertical lines as before, making 
them equal in length to the similar lines in G. The vertical line 
1U-11 is a duplicate of a-11 . Now, connect each of these points 
by dotted lines, as shown, thus obtaining the true lengths of 
similar dotted lines in the elevation. Having obtained the true 
lengths of the solid and dotted lines in the elevation, the pattern 
is now laid out in precisely the same manner as the patterns for 
problems developed by the regular method of triangulation ex¬ 
plained in the previous chapter. To develop the full pattern, 
shown at D, first draw the center line 7-8, making it equal in 
length to the line 7-8 in the elevation, which is shown in its true 
length. 

With 8 as center and radius equal to 8-9 of the half profile C, 
describe small arcs, which intersect with another struck from 7 
as center, with a radius equal to the dotted line 7-9 in diagram 
H, thus establishing the position of the two points numbered 9 
in the lower edge of the pattern. From 7 as center, with a ra¬ 
dius equal to 7-6 of the half profile B of the upper base, strike 





TRIANGULATION, SIMPLIFIED METHOD 


105 



























106 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


small arcs, which intersect with other arcs struck from points 9 
as centers, with a radius equal to the solid line 9-6 in diagram G, 
thus locating the position of point 6 in the upper edge of the 
pattern. Continue using the true lengths of the solid and 
dotted lines in the diagrams G and H, in connection with the 
lengths of the spaces in 
the half profiles B and 
C, to develop the upper 
and lower lines of the 
pattern, using each com¬ 
bination alternately un¬ 
til the pattern is com¬ 
pleted. As each point 
on the pattern is locat¬ 
ed, it should be num¬ 
bered, and the solid and 
dotted lines may be 
drawn across the pat¬ 
tern, as shown. These 
lines are not necessary, 
as each point is simply 
used as a center from 
which to find the next 
point beyond. No mat¬ 
ter what shape the pro¬ 
files of the ends of the 
article may be, the above method of development can be used in 
every case where the two halves of the article are symmetrical. 


Problem 81. Roof Collar, Round Top 
and Square Base. 



Problem 81. Roof Collar Having Round Top and 
Square Base 

In Figure 118, 1-7-8'-9' is the elevation of a roof collar having 
a round top and square base to fit over a slanting roof, indicated 


by the line ran. A shows the elevation, C the half profile of the 
top, and B the half profile of the base. The drawing is made to 
a scale of 4 inches to the foot. Draw the elevation and half pro¬ 
files of the upper and lower base. 

Find the true lengths of the solid lines in the elevation by the 
method explained in the previous problem, and develop a full 
pattern, placing the seam on the long side of the collar, as shown 
by the line 7-8 in the elevation. 

Problem 82. Three-Piece Offset Fitting 

Figure 119 shows the method employed when offset pieces 
and transition elbows are developed in three pieces, and the prin¬ 
ciples given in Problem 80 can be applied to any tapering transi¬ 
tion piece, no matter what profile either end may have. 

Figure 119 shows the elevation of a three-piece tapering off¬ 
set fitting, a portion of a round pipe joining it above and below. 
B is the upper arm with half profile, shown by A, and G the 
lower arm, the half profile of which is shown at F. The miter 
lines l'-7 f and 11*-8' are not found by bisecting the exterior 
angles, but may be established at pleasure. 

The half profiles A and F are divided into the same number of 
equal spaces and from these points draw vertical lines parallel to 
the upper and lower arms until they intersect the miter lines 1 '-7' 
and 1 U'-8'. A one-half pattern for the upper arm B and the lower 
armG is shown at H and D. These are obtained by the parallel 
method as explained in Problem 1 on the development for a two- 
piece elbow in a round pipe. The pattern for the middle sec¬ 
tion C forms a transition piece and is developed by triangulation 
as follows: 

Draw solid lines in C, connecting the points on the miter lines, 
as 2'-13', 3'-12', U'-l 1', etc.; then draw diagonal dotted lines, as 
shown. Next, obtain the true lengths of the solid and dotted 






TRIANGULATION, SIMPLIFIED METHOD 


107 

















































108 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


lines in C by taking those distances and placing them on the 
horizontal lines in diagrams 0 and P, as shown by similar num¬ 
bers, and from these points erect vertical lines, making them 

equal in length to simi¬ 
larly numbered lines in 
the half profile A, which 
are measured from the 
center line 1-7. Upon 
the end of the horizon¬ 
tal line in 0 and P, 
erect a perpendicular, 
as 14~H and A-ll. 
Upon these lines set off 
from 14 and A the sev¬ 
eral distances of the 
points in the hah pro¬ 
file F, measuring in each 
case from the center line 
14-8 of the lower base, 
as indicated by the fig¬ 
ures 11, 10-12, 9-13. 

Problem 82. Three-Piece Offset Fitting. Now, connect these 

points with points 1, 2, 
3, 4, etc., on the vertical lines previously drawn, by means of 
solid and dotted lines, as shown. These lines will then repre¬ 
sent the true lengths of similarly numbered solid and dotted 
lines on middle section C in the elevation. After the true 
lengths have been obtained, the pattern for C is developed in ex¬ 
actly the same manner as explained in connection with devel- 
ing the pattern D in Figure 117; the only difference is that the 
spaces on the upper and lower edge of the full pattern shown at 
R are not equal in length and the various distances are taken 


from the miter cuts on the half patterns H and D. Take the 
distance l'-14' in C, which shows its true length, and place it on 
the center line of the pattern, as shown by 1-11* in R. Using 
l'-2' in half pattern H as radius, and 1 in pattern R as center, 
describe arcs 2 and 2, which intersect by arcs struck from 14. as 
center with a radius equal to the dotted line 14~2 in diagram 0. 

Now, with 14'-13' in 
half pattern D as radius 
and 14 in R as center, 
describe the arcs 13 and 
13, which intersect by 
arcs, struck from 2 and 
2 as centers, with a ra¬ 
dius equal to the solid 
line 2-13 in diagram P. 
Proceed in this manner, 
using alternately as 
radius, first, the divi¬ 
sions in the miter cut in 
pattern H, then the true 
lengths of the dotted 
lines in diagram O'; the 
divisions in the miter 

cut in pattern D, then 
Problem 83. Three-Piece Reducing Elbow. the true lengths of the 

solid lines in diagram P. The edge lines 7-8 in the pattern are 
shown in their true lengths by the line 7'-8' in the elevation. 

Problem 83. Three-Piece Reducing Elbow, Round to 
Round 

Figure 120 presents the elevation, plan and half patterns for a 
three-piece reducing elbow, and shows the method of proce- 








TRIANGULATION, SIMPLIFIED METHOD 


109 

































































110 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


dure when the halves are symmetrical, and the patterns are de¬ 
veloped without the aid of a plan. In this problem, the upper 
and lower pieces, shown by C and B, are developed by the 
parallel-line method, and the middle piece A by triangulation. 
First, draw the elevation and place a half profile at the end of C 
and the half profile at the end of B. Divide both profiles into an 
equal number of spaces, as shown by the figure 1 to 7 and 8 to lk. 

From the divisions &-to lk in the lower half 
profile, draw vertical lines until they intersect 
the miter line from 1 k' to 8', and develop the 
half pattern for B, as shown at F. In cor¬ 
responding manner, from the points 1 to 7 
in the half profile of the upper piece C, draw 
horizontal lines which intersect the miter 
line from points 1' to 7', and develop the 
pattern for the upper arm C in the usual 
manner, as shown at G. Now, connect the 
points on the miter lines by means of solid 
and dotted lines. The solid and dotted lines 
in middle piece A will then represent the 
bases of sections to be constructed, whose 
altitudes are equal to the various heights in 
the half profiles. The true lengths of the solid 
and dotted lines in C are shown by correspond¬ 
ingly numbered lines in diagrams D and H, 
and are obtained by the method described in 
Problem 82. The half pattern for the middle piece, shown at P, 
is also laid out as described in the previous problem. The divi¬ 
sions between lk and 8 in the half pattern P are obtained from 
the divisions along the miter cut in the half pattern F, and the 
divisions from I to 7 in half pattern P are obtained from the di¬ 
visions along the miter cut in the half pattern for C, shown at C. 


Problem 84. Three-Piece Elbow, Oblong 
to Round 

In Figure 121 is shown the side elevation of a three-piece 
transition elbow, oblong to round in form. A half profile of the 
round horizontal arm A is shown at F, while the half profile of 
the oblong vertical arm G is shown at C. The middle section P 
is a transitive piece necessary to form a connection between the 
upper and lower arms, and is developed by 
triangulation. The patterns for both the 
upper arm A and lower arm G are developed 
by the parallel-line method, and the condi¬ 
tions given in this problem are essentially the 
same as those of Problem 83. It makes no 
difference whether the transition piece is 
shaped as shown in Figure 120, or as shown 
in this problem, the same methods apply 
in each case. 

Problem 85. Three-Piece Transition 
Elbow, Round to Square 

The principles of the simplified method of 
triangulation may be applied to any three- 
piece elbow whose halves are symmetrical, 
without respect to the shape of the end 
pieces or the angle cf the elbow. Should an 
instance arise where the halves as viewed in 
plan are not alike, a plan view must necessarily be used. Fig¬ 
ure 122 shows the side elevation of a three-piece elbow, 
of which the lower arm A is round, and the upper arm C is 
square. The middle section B is a transition piece from round 
to square, and is developed by triangulation. 

Draw the elevation and place the profile G on the end of C and 



Problem 84. Three-Piece Elbow, 
Oblong to Round. 




TRIANGULATION, SIMPLIFIED METHOD 


111 

















































112 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


profile F in its proper position below A. Develop the patterns 
for the end pieces A and C by the parallel-line method, and the 
pattern for the transition piece B by the simplified method of 
triangulation, as described in the preceding problems in this 
chapter. 

Problem 86. Furnace Boot, Round to Rectangular 

Figure 123 shows the method of developing the pattern for an 
angular furnace boot, round to rectangular. First, draw the 
side elevation of the boot, as shown, and above the upper end 
place the profile of the rectangular pipe, as shown at A. In its 

proper position on 
the lower end, draw 
the semi-circle repre¬ 
senting the half pro¬ 
file of the round pipe, 
shown at B. Divide 
the semi-circle into a 
number of equal 
parts, as shown from 
1 to 7. From these 
points on the half 
profile, draw lines at 
right angles to the 
center line 1-7 inter¬ 
secting the line l'-7' 
at 2', 3', V, 5' and 6'. 

From 6', 5' and 1+ 
draw lines to the up¬ 
per corner 8, and from 1+', 3' and 2' draw lines to corner 9. The 
' next step is to find the true lengths of the various lines in the 
side elevation, and it will be necessary to construct the dia¬ 


grams, shown at C and G, by the method explained in the 
previous problems. To find the true length of the line l+'-8 
in the side elevation, take this distance and place it on the 

horizontal 
line 8-g, as 
shown from 
8 to 1+ in dia- 
gramC. From 
8 and 4'draw 
perpendicu- 
lar lines, 
making 8-m 
equal to m-8 
in profile A, 
and 1 +' - 1 + 
equal to the 
distance from 
the line 1-7 
to point 1+ in 
the half pro¬ 
file B. Then 
m-U in dia¬ 
gram C is the true length of the line 8-1+' in the side elevation, 
and all the true lengths are obtained in the same manner. 
The first step to construct the pattern H is to draw the center 
line 8-7 equal in length to 8-7' in the elevation, which is 
shown in its true length. At right angles to 7-8 draw the line 
m-m' equal to the width of profile A, and draw lines from m 
and m' to 7 in pattern H. 

With radii equal to m-6, m-5 and m-J+ in diagram C, and with 
m and m' as centers, describe the small arcs, 6, 5 and 1+. Now 
set the dividers equal to the spaces contained in the half profile 











TRIANGULATION, SIMPLIFIED METHOD 


113 


3 














































114 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


B in the elevation, and, starting at 7 in pattern H, step from 6, 
one arc to another, and draw lines from m and m' to 4 - With 
mn in profile A as radius, and with m and m' in pattern H as 
centers, describe the arcs n and n', which are intersected by arcs 
struck from 4 as center and n-l+ in diagram G as radius. Now 
with n and n' as centers, and with n-3, n-2 and n-1 in diagram G 
as radii, describe the arcs 3, 2 and 1 . Set 
the dividers equal to the spaces 4-3, 3-2 
and 2-1 in the semi-circular profile, and, 
starting at point 4 in the pattern, step to 
arcs 3, 2 and 1 , respectively. Draw a line 
from 1 to n and n to 4 * Now, with radius 
equal to 9-1' in the elevation, which shows 
its true length, and 1 in pattern as center, 
describe the arc 9, which is intersected 
by an arc struck from n as center and 9-n 
in the rectangular profile A as radius. 

Complete the pattern by tracing a curved 
line thru the various points, and to the 
upper edge add the vertical collar, shown 
by a-b-8-9, in the side elevation. The 
lower collar, shown by l'-C-7'-e, is 
simply a straight piece of round pipe, 
for which a pattern is not required. 

Problem 87. Two-Piece Tapering Y-Joint 

There are many problems arising in the course of the sheet- 
metal worker’s experience where it is necessary to take from a 
leader or main pipe, several branches, each of which shall run in 
a different direction, and the fitting should be so constructed as 
to provide an easy flow for the contents of the pipe. 

Figure 124 illustrates the principles for developing a Y-joint, 


consisting of two tapering pipes joining a larger pipe at an angle, 
and is a fitting that is generally recognized by sheet-metal work¬ 
ers as well adapted to work of this kind. The elevation of the 
larger pipe is shown by a, b, c, 8, and is simply a straight piece of 
pipe with which the branches of the Y are to be joined at its 
upper end, shown by the line a-8. Bisect the line a-8, and from 
this point, as at 11', draw the line 11'- 4 
at the given angle to serve as the center 
line of prong F. At any convenient point 
on this line, as 4', erect the perpendicular 
l'-7', making it equal in length to the 
diameter of the small end of the tapering 
branch. Next, draw the vertical center 
line U-11', and connect 11^-1' and 7'-8, 
completing the elevation of branch F. 

The center line 1U-11' represents the 
seam line or miter line that separates the 
two branches of the fitting, and the drafts¬ 
man must select an outline that shall be 
adapted for a full view of a section on this 
line. The usual practice is to draw an 
irregular curve from 14 to a; then, ll^-a- 
11' will represent a half section on 11>-11'. 
With 11' as center and 11'-8 as radius, 
describe the quarter circle 8-11, which will represent a quar¬ 
ter section of the large pipe, as shown at C. Now, draw 
the half profile A and divide the semi-circle into a number 
of equal parts, as shown from 1 to 7. From these points 
and at right angles to the center line 1-7, draw lines which 
intersect the line l'-7', shown by 1,2', 3', 4 ', etc., on the 
upper end of branch. Since a number of points are to be 
located on the miter line and the lower base of the branch, these 



Problem 87. Two-Piece Tapering Y-Joint. 




TRIANGULATION, SIMPLIFIED METHOD 


115 


points should be equal in number to those on the upper end. 

An inspection of the drawing will show that of the six spaces 
required for the lower end of branch F, three of them are located 
on the outline of the quarter section, shown by the points 9 and 
lOatC, and the remaining three on the outline of the half section 
on the miter line, 
shown by the points 
12 and 13. From the 
points 12 and 13, draw 
horizontal lines, which 
intersect the miter 
line at 12' and 13', and 
from points 9 and 10, 
draw vertical lines to 
the base where they 
are designated by the 
numbers 9'-l O'. Con¬ 
nect the various points 
on the upper and lower 
end of the branch by 
solid and dotted lines 

drawn in the usual p ro biem 88. Irregular Two-Branch Y-Joint. 
manner. 

The true lengths of these lines are shown in diagrams G and H 
and are obtained by the method previously shown for such cases. 
For example: To obtain the true length of the dotted line 
113' in F, place the length of that line on the horizontal line in 
diagram G, shown from 8 to 3'; erect a perpendicular on each end 
of this line, making 3'-3 equal to 3'-3 in half profile A, and 8-11 
equal to 11'-11 in half section C; draw a line from 11 to 3 in 
diagram G, which is the true length of the dotted line ll’-3' 
in F. The true lengths of all the solid and dotted lines are found 


in a similar manner. As both branches F and D are similar, the 
pattern for one branch only is required. 

Placing the seam on the short side of the branch, as 1 '- 1 4 in F, 
the pattern is developed as follows: Draw the center line 7-8 
in pattern R equal in length to 7'-8 in the elevation. With 7-6 
in half profile A as radius and 7 in R as center, describe the arc 6, 
which intersect by an arc struck from 8 as center and 8-6 in 
diagram G as radius. Then, with 8-9 in quarter section C as 
radius and 8 in R as center, describe the arc 9, which intersect 
by an arc struck from 6 as center and 6-9 in diagram H as 
radius. 

Proceed in this manner until the line U~H in the pattern is 
obtained. The divisions 12,13,1 bin the pattern are then ob¬ 
tained from half section B. Then, using k-3 in half profile A as 
radius and U in pattern R as center, describe the arc 3, which 
intersect by its proper radius found in G. With 11 in the pattern 
as center, and a-12 in half section B as radius, describe the arc 12, 
which intersect by an arc struck from 3 as center, and 3-12 in 
diagram H as radius; then use the spaces in the half sections A 
and B, and the true lengths in diagrams G and H, until the line 
1-1U is drawn, which is shown in its true length from 1' to H 
in F. A line traced thru the various points completes the pat¬ 
tern. Laps should be allowed for seaming together, as well as 
for seaming to the collars which are placed at each opening. 

Problem 88. Irregular Two-Branch Y-Joint 

Figure 125 shows the elevation and half sections of an irregu¬ 
lar two-branch Y-joint, which is drawn to a scale of 3 inches 
to the foot. The main pipe, shown by mneh, is round in form, 
the opening in the upper end of branch B is square, and the 
opening in branch A is round. 

The half circle F represents the half section on the base line 







116 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


mn, G the half section on the line bd, and C the half section of the 
square pipe on the line ug. H in the side elevation shows the 
half section on the miter line 7-a. Draw the elevation and sec¬ 
tions, as shown, and develop the pattern for each branch by the 
method explained in the previous problem. 

Problem 89. Two-Branch Y-Joint, Round to Round 

Figure 126. This figure is drawn to a scale of 3 inches to the 
foot, and shows the elevation and sections of a two-branch 
tapering Y similar in form to Figure 124. The branches A and 
B are mitered to vertical pipes F and G. Draw the elevation of 

the prongs; also the 
half profiles of the 
vertical pipes and 
the quarter section 
C of the large pipe. 
The height of the 
miter line between 
the two prongs at 
lly-a is made equal 
to the semi-diame¬ 
ter ma of the large 
pipe. In other 
words, the true half 
section R on the mi¬ 
ter line 1 Jf-a consti¬ 
tutes a quarter circle 
with a radius equal 
to one-half the di¬ 
ameter of the large 
pipe. Having drawn the half profile H and the sections R and 
C, find the true lengths of the solid and dotted lines and develop 


the pattern for branch A in the same manner as described in 
connection with branch F in Figure 124. The divisions between 
1 ’ and 7' on the upper edge of the pattern are obtained from the 
divisions along the miter cut of the pattern for the vertical pipe 
G, which is developed by the parallel-line method, the stretch¬ 
out line being placed in the position shown by the line be. 

Problem 90. Two-Branch Y-Joint, Round to Square 

The problem pre¬ 
sented in Figure 127 
is similar to that 
described in the 
previous problem, 
with the difference 
that the upper end 
of both branches are 
mitered to vertical 
pipes A and H, 
which are square in 
form; the half pro¬ 
files are shown at B 
and R. The true 
half section on the 
miter line 7-a is 
shown at F, and is 
obtained in the same manner as in Figure 124. The quar¬ 
ter circle n-U represents a quarter section of the round pipe, 
shown atG. Figure 127 is drawn one-third full size, and, when 
enlarging this and the various problems in this chapter, more 
spaces should be used in dividing the profiles. 

Draw the elevation, half profiles and sections, as shown, and 
develop the patterns for branch C and the vertical square pipe 









TRIANGULATION, SIMPLIFIED METHOD 


117 



















































118 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


A. As both branches C and D are similar, only one pattern is 
required. 

Problem 91. Irregular Two-Branch Fitting, Round to 
Round 

Figure 128 shows the side view and sections of a two-branch 
fitting which is drawn 
to a scale of 3 inches 
to the foot. The 
main pipe and 
branches are round 
in form, the half pro¬ 
files of which are 
shown by the semi¬ 
circles C, D and H. 

Branch A is in a ver¬ 
tical position, while 
branch B is inclined 
at an angle of 45° to 
the base line mn. The 
miter line between 
the branches A and B 
is obtained by draw¬ 
ing a line from the 
intersection of the 
two branches at point 10, to the center of the upper base of 
the large pipe, shown at a. The half section on the miter line 
10-a is shown atG, and is drawn as follows: 

From point a at the lower end of the miter line, and at right 
angles to 10-a, draw the line a-8, making it equal in length to the 
semi-diameter of the large pipe, as shown by a-8 in half profile H. 
Connect points 10 and 8 by means of an irregular curve. Then, 
10-8-a will represent a half section on miter line 10-a. Draw the 


elevation and sections, as shown, and develop separate patterns 
for branches A and B by the method described in Problem 87. 

Problem 92. Irregular Fitting, Round to Square, Miter¬ 
ing with Vertical and Horizontal Connections 

In figure 129, ABRCF represents the side elevation of an 
irregular Y fitting, the base R being round, the branch C miter¬ 
ing with the horizontal square pipe F, and branch B mitering 
with the vertical round pipe A. The circle D represents the 
section on the line ab, H the section of the square pipe on the 
line Ce, and the semi-circle R the half section of the large pipe 
on the line 6'-l2'. The height of the miter line 15-9' is made to 
equal 6'-9', or the semi-diameter of the large pipe R, and this 
distance is taken as a radius for describing the quarter circle 
which represents a true half section on the miter line 15-9', 
shown at G. 

Draw the elevation and place the profiles and sections in 
position as shown. Space the profile D into a number of equal 
divisions, and from these points draw vertical lines, which inter¬ 
sect the miter line from 1 ' to 5'. At right angles to branch A, 
draw the stretchout line 5-5, upon which place the girth of pro¬ 
file D, as shown from 5 to I to 5. Draw the measuring lines 
which intersect by horizontal lines draw from similarly num¬ 
bered points on the miter line 1 '-5'. A line traced thru these 
points will give the miter cut of the full pattern for branch A, 
shown at J. The divisions on the miter cut in pattern J are 
used in developing the upper edge line of the pattern for branch 
B in the elevation. The half profile R of the large pipe is now 
divided into equal parts, as shown from 6' to 12', and from these 
divisions vertical lines are drawn to the base line in elevation. 
In corresponding manner, space the half section G, as shown 
from 13 to 15, and draw horizontal lines which intersect the 








TRIANGULATION, SIMPLIFIED METHOD 


119 

















































































120 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


miter line 15-9', as shown. Now, connect the points on the 
upper and lower bases of the transition piece B by means of solid 
and dotted lines. The true lengths of the solid lines are shown 
in diagram M, and those for the dotted lines in diagram N, and 
are found by the 
method given in pre¬ 
ceding problems. 

The half pattern 
for branch B is shown 
at P. The true 
lengths of the solid 
and dotted lines are 
obtained from dia¬ 
grams M and N, re¬ 
spectively. The 
spaces from 6 to 9 to 
15 on the lower edge 
of the pattern are 
obtained from divi¬ 
sions of correspond¬ 
ing numbers in sec¬ 
tions G and R, respectively. The divisions from 1 to 5 on the 
upper edge of pattern P are obtained from the divisions along 
the miter cut in pattern J. The same method is employed for 
developing the patterns for transition piece C and horizontal 
branch F. 

The half pattern for C is shown at E; the divisions 12 to 9 and 
9 to 15 on the lower edge of the pattern are obtained from simi¬ 
larly numbered spaces on half section G and half section R of the 
large pipe. The true lengths of the lines in pattern E are shown 
in diagrams 0 and L ; the vertical lines 17-X and 16- X are equal 
in length to mg or na in profile H. The full pattern for branch 


F is shown at T, and is developed by the parallel method in the 
usual manner. 

Problem 93. Three-Branch Fitting 

Figure 130 shows the method employed when developing pat¬ 
terns for fittings which contain three or more branches, and is 
also applicable, whether the branches are inclined at the same 
angle, or whether the openings in the branches are of the same 
shape or not. In this problem we have three branches, all in¬ 
clined at the same angle and whose openings at the upper end 
are of the same diameter, so that the pattern for one branch is all 
that is required. The three branches are to be carried off from 
a 5-1/4-inch pipe in such manner that the angle made between 
the center line of each branch and the axis of the main pipe is 
one of 133°. The open ends of the branches are to be equal, 
each being 3 inches in diameter. 

In laying out the drawing, it is only necessary that one branch 
be drawn at right angles to its center line mn in plan, as shown 
by l-lt'-7'-8-lk in the partial elevation. First, describe the 
circle J^-g-m in plan, whose diameter shall be equal to the large 
pipe. The outline of the circle is next divided into three equal 
parts, thus locating the points <7 and m, and from these points 
draw the miter lines m- 1 ', g- 1 ' and From the center of 

the circle in the plan, draw the vertical center line 1 '- 1 , making 
0-1 equal to the semi-diameter of the large pipe. Next, from 0 
on this line, draw a line at an angle of 45 °, to serve as the center 
line of the branch pipe. Locate point II' on this line, 6 inches 
from 0, and erect the perpendicular II 4 .- 8 . 

The half profile, shown at B, is constructed, and the outline of 
the semi-circle divided into a number of equal spaces. The 
points thus located are then projected to the line I U~ 8 , as shown. 
Next, thru 0 draw the line k~7' equal in length to the diameter 






f\> 0\ Gi 


TRIANGULATION, SIMPLIFIED METHOD 


121 










































122 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


of the large pipe. With 0 as center, and 0-4 as radius, describe 
the arc 4-1. Then 1 -0-4 will represent the true section on 4 '-1' 
in plan, as shown at C. Divide this section into equal parts, and 
from these divisions draw vertical lines to the miter line 1 '-4' in 
plan, intersecting this line at 2' and 3', as shown. With 1' as 
center and radii equal to 2' and 3', describe 
arcs which intersect the miter line 1 '-4, as 
shown by 2 and 3. From these points 
draw vertical lines, which are intersected 
by horizontal lines drawn from similarly 
numbered points in section C. A curved 
line traced thru these points of intersec¬ 
tion is then the foreshortened miter line, 
as shown by 1-2'-3'- 4', in the elevation, 
and by the miter line 1 -4 in plan. 

Now, divide the distance from 4 to 7 
on the arc of the large circle in the plan 
into equal parts, as shown by 4, 5, 6 and 7, 
and from these points draw vertical lines 
intersecting the base line of the fitting in 
the elevation at 4', 5', 6 ' and 7' Next, 
draw solid and dotted lines in branch A 
and obtain their true lengths, as shown in 
diagrams F and G. 

To find the true lengths of the dotted 
line 2'-12', place this distance as shown 
from 2' to 12' in diagram G, from which draw the vertical lines 
2'-2 and 12'-12, making 2'-2 equal to the distance measured 
from the point a on the line mn in plan to the point 2 on the 
miter line 4 -1 and the line 12'-l 2 equal to 12'-l 2 in half profile 
7? in elevation. The distance from 2 to 12 in diagram G is 
then the true length of the dotted line 2'-12' in the elevation. 


The true lengths of all the solid and dotted lines are obtained in 
similar manner, and the pattern developed by the method 
described in Problem 87. 

The full pattern for branch A is shown in R. The divisions 
from 8 to 74 on the upper edge of the pattern are obtained from 
the half profile in B. The divisions from 
7 to 4 are equal to the divisions from 7 to 
4 on the outline of the circle, which repre¬ 
sents the large pipe in plan. The divisions 
from 4 to I in pattern R are taken from 4 
to 1 in section C. 

Problem 94. Three-Way Branch 

In Figure 131 is shown the elevation 
patterns for a three-way branch fitting, 
and round to round. The branches A, B 
and C are carried off from a 5-1/4-inch 
round pipe in such manner that the cen¬ 
ters of the round openings at the upper 
end are in the same vertical plane, thus 
making both sides of the branch symmet¬ 
rical. First, draw the center line 18-11. 
From point 11' on this line draw the cen¬ 
ter line of the branch B at an angle of 60°, 
making it 8 inches long, as shown by 11'- 
4'. At right angles to the center line of 
branches A and B, draw the lines 1-7 and a-15, making them 
3 inches long, or equal to the diameter of the outlet of each 
branch. 

Next, draw the base line m-8 equal to the diameter of the 
main pipe, and place the semi-profiles in their proper position, as 
shown by the semi-circles G, H and F. The next step is to 



Problem 94. Three-Way Branch. 









TRIANGULATION, SIMPLIFIED METHOD 


123 

















































124 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


obtain a true section on the miter line 1 k-11', which is constructed 
in the following manner: 

From point 11' at the lower end of the miter line and at right 
angles to 11'-lk, draw the line 11'-b, making it equal to the 
semi-diameter of 
the large pipe, as 
shown by 11'-8 in 
profile F. Connect 
points lk and b by 
means of an irregu¬ 
lar curve, which 
will represent a true 
half section on the 
miter line lk-11', 
shown in R. As the 
side branches B and 
C are alike, only one 
pattern will be re¬ 
quired ; also a sepa¬ 
rate pattern for the 
center branch A, 
both of which will 
be developed by triangulation. To develop the pattern for 
branch B, first divide the quarter section F and half section R 
into three equal parts, as shown by the figures 8 to 11 and 11 to 
lk', then divide half profile H into six equal parts, as shown 
from 1 to 7. From these points, at right angles to the various 
base lines, draw lines intersecting the base lines, as shown 
by similar numbers. 

Draw solid and dotted lines in branch B and find their true 
lengths by constructing the diagram, shown at E and J, by the 
method explained in previous problems. The full pattern for 


branch B is shown in M . The divisions from 7 to 1 on the upper 
edge of the pattern are obtained from the half profile in H. The 
divisions from 8 to 11 are equal to the divisions from 8 to 11 on 
quarter section F, and the divisions 11 to lk are equal to the 
spaces on half section R. As the four quarters of center branch 
A are alike, divide one-half of semi-profile G into three equal 
parts, and from these divisions draw vertical lines to the center 
line a-15, as shown by 17' and 16'. From these intersections, 
draw solid and dotted lines to points 12' and 13' on miter lines 
lk-H'. The true lengths of these lines are shown in diagrams 
P and L. 

A one-half pattern for branch A is shown in N. The divisions 
from 11 to lk are equal to the divisions from 11 to Ikon half sec¬ 
tion R. The divisions on the upper edge of pattern N are equal 
to the divisions from 18 to 15 in semi-profile G. The patterns M 
and N are developed in precisely the same manner as described 
in connection with Figure 124. 

Problem 95. Two-Way Transitional Branch Mitering with 
Horizontal and Vertical Connections 

Figure 132 presents the shortened method employed in devel¬ 
oping the patterns for a two-way transition branch. The main 
pipe, shown by e-g-15-1 in the elevation, is round in form. The 
opening in branch G is rectangular in shape and is mitered to a 
horizontal pipe. Branch F is tapering in form, the upper end 
being mitered to the round vertical pipe, shown by a-b-lk-8 in 
the elevation. 

First, draw the elevation of the tapering branch F, and de¬ 
scribe the semi-circles D and H, which represent the half profiles 
of the upper and lower bases. Divide profiles D and H into the 
same number of equal divisions, and project these points to the 
upper and lower base lines in the usual manner. Connect these 



Problem 95. Two-Way Transitional Branch, 
Horizontal and Vertical Connections. 






TRIANGULATION, SIMPLIFIED METHOD 


125 













































































126 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


points by solid and dotted lines, as shown by 10'-3', 9'-2 f , etc., 
in the elevation. Next, from point 4' on the lower base line of 
branch F, draw the center line >/-t of branch G at an angle of 50 ° 
to the base line, and draw the outline of branch G, as shown by 
4'-7-V- X-l . Draw the profile of the rectangular pipe, shown 
by A, B, C and D; also the half section on the joint or miter line 
7-^', shown at R. 

From points 5' and 
6’ on the miter line, 
and 3' and 2' on the 
lower base line, draw 
lines to the corners 
A' and C" in branch 
G, as shown. The 
true lengths of these 
lines are shown in 
diagram T, the dis¬ 
tance m'c" being 
equal to one-half 
the width of the 
rectangular pipe, 
shown by mA and 
nC m the profile. 

The true lengths 
of the solid and dot¬ 
ted lines in the tapering branch F are shown in diagrams P and 
L, and a one-half pattern for same is shown in E. The spaces 
from 14 to 8 on the upper edge of the pattern E are equal to 
the spaces from 14 to 8 in half profile D. The spaces from 15 
to 4 on the lower edge are equal to the spaces from 15 to 4 on 
the half profile of the main pipe, shown in H. The spaces from 
4 to 7 are equal to the spaces from 4" to 7 on half section R. 


A one-half pattern for transitional branch G is shown in J; the 
divisions from M to n on the upper edge are taken from profile 
0. The divisions from I to 4 and 4 to 7 are equal to correspond¬ 
ingly numbered divisions on half profile H and half section R. 
The full pattern for the horizontal rectangular branch is shown 
at K, and is developed by the parallel-line method. 

Problem 96. Irregular T-Joint, Rectangular 
to Round 

Figure 133 shows the simplified method employed in devel¬ 
oping the patterns for an irregular T-joint, changing from a 
round to a rectangular outlet. A plan view is here shown, but 
is not required for the development of the pattern, as both 
valves are symmetrical. 

First, draw the side elevation in which BDF H represents the 
round pipe. From point 4' on the center line, draw the line 
4 '-C, which represents the angle of the T on its center line, and 
construct the elevation of the transition piece, as shown by 
4 ’-V-G'-A'-l'. 

The section or profile of the rectangular pipe on the line 
G'-A’ is indicated by SFRA in profile#, thru which center line 
ab is drawn. Next, draw half profile H of the round pipe. As 
both quarter circles are symmetrical, it is only necessary to 
divide the quarter circle into an equal number of spaces, as 
shown by the figures 1 to 4, and from these points draw vertical 
lines, which intersect the miter lines in the elevation at points 
marked 1' to 4' and 4' to 1". From the intersections on the 
miter line 4' to 1", draw lines toG', and from the divisions on the 
miter line 4' to 1 ', draw lines to A'. These lines then represent 
the bases of sections having altitudes equal to the heights in the 
semi-profiles E and H, and their true lengths are found by con¬ 
structing the diagrams of triangles, shown at R and L. For ex- 



Problera 96. Irregular T-Joint, Rectangular 
to Round. 




TRIANGULATION, SIMPLIFIED METHOD 


127 




































































128 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


ample: To find the true length of the line U'-G' in the elevation, 
take this distance and place it on any line as a-p in diagram L; 
draw the perpendicular a'-G and U'-k equal in height to S-a in 
profile E, and 4-m in profile H. A line drawn from G to 4 in the 
diagram will be the true length of the lineG'-V in the elevation. 
Again, for an example: To find the true length of the line A '-S' 
in the elevation, take this distance 
and place it on the base line in dia¬ 
gram R, as shown from b to S'. 

Draw the vertical lines b-A and 3'- 
3, equal, respectively, to Ab in pro¬ 
file E, and 3-n in semi-profile H. 

A line drawn from 3 to A will be 
the true length of this line. In 
this manner, all of the true lengths 
are obtained. Diagram R shows 
the true lengths of the lines in J, 
and diagram L the true lengths of 
the lines in P. 

The pattern for the opening in 
the vertical pipe is shown in T, from 
which the stretchout of the lower 
edge of the pattern for the transi¬ 
tion piece can be obtained. To obtain the pattern for the 
opening in the vertical pipe, take the stretchout of the semi¬ 
profile H and place it on the line eg, as shown by the numbers 
.4 to 1 to k- From these divisions, draw vertical lines, which are 
intersected by horizontal lines drawn from similarly numbered 
intersections on the miter line in the elevation. A line traced 
thru these points of intersection will give the pattern for the 
opening to be cut in the round pipe to receive the transition 
piece. The full pattern for the transition piece is shown in K. 


The distance A’-l’ and a-1 are equal to G'-l" and A'-l' in the 
elevation, which are shown in their true lengths. The distances 
aG, GA, AR, etc., on the upper edge of pattern K are ob¬ 
tained from the profile of the rectangular pipe, shown at E. 
The true lengths from the diagrams R and L, and the divisions 
on the lower edge of the pattern are obtained from similarly 
numbered divisions in the pattern 
for the opening in the vertical pipe, 
as shown in T. 

Problem 97. Two-Way Elbow 
Fitting 

The sheet-metal worker engaged 
in the construction of fittings and 
transition pieces in pipes' used for 
exhaust and heating systems has 
frequent use for a two-way elbow 
fitting, shown in Figure 134. The 
two branches are made in the form 
of five-piece elbows which are con¬ 
nected to the main pipe by a transi¬ 
tion piece, shown by m-A-B-1 in 
the elevation. The arrangement of 
the fitting is such that the elbow sections C and C' intersect 
each other for a certain distance on the vertical center line, as 
shown by the line ag. 

First, draw the elevation of the elbows in correct position, the 
distance m-1 being made equal to the diameter of the main pipe. 
Next, draw the lateral outlines of the transition piece which are 
represented by the vertical lines mA and B-l in the elevation. 
The line AB is then drawn, completing the elevation of the tran¬ 
sition piece, to which is added a straight collar 2 inches wide, 



Problem 97. Two-Way Elbow Fitting. 




TRIANGULATION, SIMPLIFIED METHOD 


129 
















































130 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


shown by A-e-n-B. As both elbows are of equal diameter, the 
pattern for section C is the only pattern required; the patterns 
for the entire elbow may be derived therefrom. The center 
line of section C is drawn and extended to the upper portion of 
the drawing; next, on this line construct the half profile of the 
elbow, shown by the semi-circle at F. Divide the semi-circle 
into equal spaces, and from these points, draw lines parallel to 
the center line across the elevation of the elbow until they inter¬ 
sect the miter lines ga and a-1 in section C. 

The pattern for section C may now be developed by setting off 
the stretchout on the lineG H in the usual manner, and the point 
a on the miter line is located at the point a' on the half profile at 
F, and also at a" on the stretchout lineG H in the pattern. The 
upper line of this pattern will serve for the remaining sections of 
the elbow. The pattern for section G of the elbow should al¬ 
ways be developed first, because the distances between spaces 
on the lower miter cut of the pattern will be required in obtain¬ 
ing the length along the upper edge of the transition piece in its 
development. 

The transition piece is shown in the elevation by A-B-l-a-n. 
From the half plan, shown below the elevation, it will be seen 
that the lower base of this piece has for its outline a circle, shown 
by h-10-b, and that its upper base on the lines ma and a-1 are 
parts of two circles that represent the lower ends of the elbow 
sections C and C', whose position upon the plan is found in the 
following manner: 

From each of the points on the miter line a-1 in the elevation, 
draw vertical lines to the plan, crossing the center line hb. 
Measuring from line hb, each line is made equal to the distances 
from the center line 1-13 of half profile/ 1 to the points 1,3, 5, 7, 
etc., on the semi-circle. A line traced thru these points will 
show the outline of the upper base in plan. Next, the outline of 


the lower base in the plan is divided into the same number of 
equal parts. 

The triangles may now be indicated in the plan by drawing 
lines between successive points on the upper and lower bases, as 
2-3, 3-U, U-5, 5-6, etc., and afterwards projecting these lines to 
the elevation. 

The triangles on the surface of the transition piece in the 
elevation are necessary only to assist in seeing some of the ele¬ 
ments which appear somewhat confused, as seen upon the plan. 

The true lengths are found in the customary manner, as shown 
by the diagram of triangles in R and E. 

A one-half pattern of the transition piece is shown at J, and 
the method of laying out the pattern is similar to preceding de¬ 
velopments. It is necessary merely to note that the distances 
a-9, 9-7, 7-5, 5-3 and 3-1 on the upper edge of the pattern are 
taken from similarly numbered divisions on the miter cut of the 
elbow pattern, the distances being there shown in their true 
lengths. 

The divisions from 10 to b on the lower edge of pattern J are 
equal to the spaces from 10 to b in the half plan. 

Problem 98. Ship Ventilator with Round Mouth 
and Base 

In Figure 135 is shown the simplified method of developing 
the patterns for a round tapering elbow. An elbow of this form 
is generally known as a ship ventilator, and the principles shown 
in this problem are applicable to any form or shape, no matter 
what the respective profiles may be at the base or top. 

Ship ventilators of this type are used on sailing vessels and 
steamships of all kinds, and are made in a great variety of forms 
and in proportions to suit the various uses and accompanying 
conditions. No rule can be given to suit all conditions, but the 


TRIANGULATION, SIMPLIFIED METHOD 


131 



iO 









































132 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


general proportions given in this problem may be taken as 
standard proportions for designing any size ship’s ventilator of 
this type. 

Figure 135 shows a round ventilator made up in four sections; 
the half profiles of the 
upper and lower open¬ 
ings are shown in the 
elevation by the semi¬ 
circles A and B. The 
size of the base or lower 
opening constitutes the 
basis for determining 
the outline of the throat 
and heel and the pro¬ 
portions of the ventila¬ 
tor. In this case, as¬ 
suming the diameter of 
the base to be 12 inches, 
draw the base line 12 
inches long, establishing 
the points a and 6. From 
point a erect a line per¬ 
pendicular to the base 
line, extending it indefi¬ 
nitely, as shown by the 
line mn. The radius of 
the throat is next determined by taking one-fourth the diameter 
of the base, or 3 inches, which is marked off on the base line ex¬ 
tended to the right from point b to point e. Using e as center 
and eb as radius, describe the quarter circle, or throat, line be, 
and intersect it by the perpendicular line erected from e at C. 
As the diameter of the mouth of the ventilator should be twice 


the diameter of the base, and the overhang one-third the diam¬ 
eter of the base, the elevation is completed inthe following man¬ 
ner: From point e on the base line, set off eg to equal one-third 
the diameter of the base, or 4 inches. From g at right angles to 
eg, draw the vertical line go. Now, with c as center and twice 
the diameter of the base, or 24 inches, as radius, describe a short 
arc, intersecting the vertical line go at h. Draw the inclined 
line he, which represents the mouth of the ventilator. 

The next step is to draw the heel of the elbow. Using a radius 
equal to one and three-fourths times the diameter of base, and 
with one leg of the dividers on point h and the line mn, describe 
short arcs intersecting at x. With x as center and xh as radius, 
draw the arch hv tangent with the vertical line mn. This 
method, in the form of a rule, shows: (1) Throat radius equals 
one-fourth diameter of base. (2) Heel radius equals one and 
three-fourths times diameter of base. (3) Mouth diameter 
equals twice diameter of base. (4) Overhang equals one-third 
diameter of base. 

The throat and heel are now divided into four equal parts, and 
these points connected by lines, which represent the miter lines 
of each section which is developed by triangulation. To obtain 
the pattern for section Number 1 in the side elevation, take a 
tracing of this section and place it at the right of the elevation, 
as shown by 7-1,1^-8 in F. 

Bisect 7-1 and 8-1U and obtain the points V and II', respec¬ 
tively. With these points as centers, describe a semi-circle on 
each side, which will represent a half profile of each end of the 
section. 

Next, divide these half profiles into the same number of parts, 
as shown by the figures I to 7 and 8 to lb. At right angles to 
1-7 from the various points on the half profile, draw lines to in¬ 
tersect the line 1-7 at 2', 3', b', 5' and 6'. Also, at right angles 



Problem 98. Ship Ventilator, Round 
Mouth and Base. 




TRIANGULATION, SIMPLIFIED METHOD 


133 

































134 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


to the base line 8-1 If. from the divisions 9,10,11,12, etc., on the 
half profile, draw lines to intersect the line 8'-lU at 9', 10', 11', 
12' and 13'. Draw solid lines from 6' to 9', 5' to 10', V toll', 3' 
to 12' and 2’ to 13'; also the dotted lines 7 to 9', 6' to 10', 5' to 
11', etc. The true lengths of the solid and dotted lines are ob¬ 
tained in the usual man¬ 
ner by constructing the 
diagram of triangles, 
shown in G and H. The 
completion of the pattern 
for section Number 1 is 
shown in P, and since the 
process is precisely similar 
to that described in pre¬ 
ceding problems that have 
been developed by the 
simplified method of tri¬ 
angulation, no further 
mention of the principles 
involved is necessary. 

A construction similar 
to that shown in F is re¬ 
quired for each section of 
the ventilator. After 
the patterns for each 
section have been devel¬ 
oped, as shown in P, 
double edges must be allowed to the lower part of the patterns 
for sections 4, 3 and 2, and single edges to the upper part of the 
patterns for sections 1, 2 and 3. In a similar manner, the pat¬ 
terns for a ship ventilator or tapering elbow, composed of any 
number of sections, may be constructed. 


Problem 99. Ship Ventilator with Round Base and 
Elliptical Mouth 

Figure 136 illustrates how the patterns are developed for a 
ship ventilator or any other form of tapering elbow, making a 
transition from one profile to another. 

The profile of the base is in the form of a circle, while the 
mouth is defined in the front elevation by an ellipse. 

First, draw the side elevation, as this view shows the outline 
and principle proportions of the ventilator. In this problem it 
may be assumed that the diameter of the base is 6 inches, and 
the overhang will equal the diameter of the base. Construct the 
right angle ABC. From the point B on the horizontal line BC, a 
distance of 6 inches is measured and the point m is located, as 
shown. The diameter of the lower opening is then measured off 
on the same line from the point m to point C. From the points 
m and C, vertical lines are drawn and the circle is then described, 
which represents the profile of the lower end of the ventilator. 
Next, from point m, the inner side of the elbow, draw the miter 
line for a five-piece elbow in the usual manner, as shown by the 
line ran. 

The lower extremity of the elliptical mouth of the ventilator is 
located by making the distance Be equal to one-fourth the diam¬ 
eter of the base. The upper point is located at A and the dis¬ 
tance eA is equal to two and three-fourths times the diameter of 
base, or 16-1/2 inches. With this distance as radius and with 
A and n as centers, describe short arcs intersecting at g. With 
the point g as center, and with gA as radius, describe the arc nA, 
which represents the heel of the ventilator. Now, with t as cen¬ 
ter and te as radius, describe the arc cm, which shows the lower 
outline of the ventilator. The arcs nA and em are next spaced 
into four equal parts, and chords are drawn between the points 
located on the outline of each arc. Complete the elevation by 





TRIANGULATION, SIMPLIFIED METHOD 


135 


drawing the miter or joint lines of each section, and bisect each 
line, locating the centers a, b, c and d. 

The next step is to draw the front elevation, altho in actual 
shop practice it is but necessary to draw a half elevation. Then 
draw any vertical line, as FG. From the points A and e in the 
side elevation, draw horizontal lines at right angles to FG, inter¬ 
secting FG at m' and n'. Take half the diameter of the round 
base and set it off from g' to g". 

Since it is customary to make the mouth of the ventilator of 
such proportions that the minor axis of the ellipse is equal in 
length to two-thirds that of the major axis, the ellipse shown in 
the front elevation may now be constructed. Take half the 
length of the minor axis and set it off from d ' to d". Having de¬ 
termined the length of the major and minor axes, the ellipse, 
shown in the front elevation, may now be constructed by 
means of the short rule described in Figure 26. Next, describe 
a graceful curve from d" to g"; these arcs form the lateral sides 
of the ventilator. 

The next step consists of bisecting each of the miter lines in 
the side elevation, shown by a, b, c, and from these points at 
right angles to FG, draw horizontal lines, which intersect the line 
FG, and cross the curve d"g" at a", b " and c". The miter lines 
in the side elevation represent the major axes, and the horizontal 
distances in the front elevation the minor axis of elliptical sec¬ 
tions to be constructed on the several miter lines in the side ele¬ 
vation. 

The pattern for lower section Number 1 is developed by the 
parallel-line method; the stretchout of this pattern is obtained 
by dividing the outline of profile H into a number of equal 
spaces, as shown. 


The patterns for sections 2, 3, 4 and 5 are developed by the 
simplified method of triangulation in precisely the same manner 
as section Number 1 of the round ventilator explained in the 
preceding problem. A partial development of section Number 
3 in the side elevation is shown in the drawing, and a similar 
course is necessary for the development of each section of the 
ventilator. Take a tracing of section Number 3 and place it, as 
shown, by r, h, s, u,mJ; altho in a different position from that 
shown in the elevation, it is merely copied same size from the 
latter view. Next, on the line su in J, which is the major axis, 
construct the semi-elliptical profile, shown in P, the minor axis 
11'-11 being taken from the front elevation, where it is 
represented by the line b'-b”. In the semi-profile R, the 
major axis rh is equal to the miter line rh in the side eleva¬ 
tion, and the one-half minor axis k'~h is equal to a'a" in the 
front elevation. 

To save time, the profiles of the upper and lower bases of each 
section may be drawn by circular arcs, as shown in Figure 26. 
Divide both profiles R and P into the same number of equal 
spaces, as shown by the figures 1 to 7 and 8 to 1U- At right 
angles to the lines rh and su, from the various divisions on pro¬ 
files R and P, draw lines intersecting rh and su, as shown. The 
surface of section J is now divided into triangles by means of 
solid and dotted lines drawn between successive points on the 
base lines rh and su. The true lengths of these lines are deter¬ 
mined by constructing a diagram of triangles, and the pattern 
developed by the method already explained in previous prob¬ 
lems in this chapter. 

Allowance of material must be made on all patterns for seam¬ 
ing or riveting. 



CHAPTER IX 

Skylights 


Skylights are known by their form as flat, double pitch, gable 
and hipped. The construction, shown in the following prob¬ 
lems, may be adapted to all forms of metal skylights. When 
designing a skylight to suit the special requirements of the par¬ 
ticular case in hand, the draftsman may encounter conditions 
that require the several parts to have profiles different from 
those given here, but the method of development will not differ 
from that given in the various problems in this chapter. 

The skylight bars must be constructed in such a manner as to 
furnish a supporting surface for the glass, and a gutter to carry 
off the water of condensation. It is also important that the 
draftsman, when designing a skylight of any form, design the 
profiles of the different parts of ample size and as simple in form 
as consistent with required strength to allow of rapid forming 
into shape on the cornice brake. 

Problem 100. Flat Skylight 

Figure 137 shows the longitudinal section and patterns for the 
most common style of flat skylights, those which are set on a 
curb projecting above the roof, to insure imperviousness to 


storms, the inclined roof having the necessary pitch. Some 
sheet-metal workers are under the impression that skylight de¬ 
velopments are very difficult. This is not so, as skylight pat¬ 
tern-drafting entails no intricate methods of development, the 
patterns being obtained by the parallel-line method, and be¬ 
comes a very simple matter if this method is thoroughly under¬ 
stood. 

A sketch of the plan view is shown at F, the miter between the 
upper and side curbs being simply a square return miter, while 
the miter between the lower and side curbs is a plain butt miter. 
The longitudinal sectional view in Figure 137 is drawn to a scale 
of 4 inches to the foot, showing the section or profile of the com¬ 
mon bar A, the lower curb B, and the upper and side curbs C. 
The common bar is of the universal type, and the upper and 
lower curbs are formed to coincide with the general dimensions 
of this bar, as shoWn. 

To obtain the pattern for the side curbs of the skylight, draw 
the stretchout line SV, and upon this line place the stretchout of 
profile C, as shown by the numbers from 1 to 10. Thru these 
points at right angles to (SV draw the usual measuring lines. 


136 









SKYLIGHTS 


137 


These are intersected by vertical lines drawn from similarly 
numbered points on the profile of side curb C. Connect the 
points of intersection thus obtained, which will give the pattern 
for the square return miter cut on the upper end of the side 
curb, laps being allowed, as shown by dotted lines. This miter 
cut without the laps is also the pattern for the miter cut of the 
upper curb, the cut of this pattern being the same at both ends. 

The miter cut on the lower end of the pattern for the side curb 
is shown at G, and is simply a 
plain butt miter and is obtained 
by drawing vertical lines from 
the points in profile B of the 
lower curb to the horizontal 
measuring lines in pattern G, as 
shown. From points 8, 9 and 10 
of profile B draw a vertical line 
which will intersect the measur¬ 
ing lines 8, 9 and 10 of pattern 
G; also from point g of profile B 
to line 7 of the pattern, as shown. 

Profile B from g to a will butt 
against the upright member of profile C from 7 to 6. The cut 
on the side curb from 7 to 6 must be a duplicate of that part of 
profile B from g or e to c. It is necessary then to take the dis¬ 
tance from e to d and place it from 7 toward 6 on the stretchout 
line SV of the pattern, as shown at d'. From this point draw 
the measuring line which is intersected by vertical lines drawn 
from d and c of profile B, giving the required cut. The rest of 
the cut of the side of profile C is straight to 5 ; then from 5 to U 
it goes back the distance cd of profile B, and is then straight 
from 4 to 3. 

The gutter of the side curb, shown by 3-2-1 in profile C, is cut 


off at an angle, as shown at xy. This is to allow the condensa¬ 
tion in the gutter of the side curb to drain into that of the lower 
curb, shown by fe in profile B. Lines projected from xy to 1-2 
in the pattern will complete the miter cut for the lower end of 
the side curb. 

A stub or short pattern of common bar A is shown atF, and is 
obtained in the following manner: At any convenient distance 
to the right and left of profiles C and B in the longitudinal sec¬ 
tion, draw the vertical measur¬ 
ing lines mn and ho, as shown. 
Next, draw the vertical line m'n' 
in pattern F, and upon this line 
place the stretchout of the com¬ 
mon bar, or profile A, as shown 
by the numbers 1 to 6 to 1 . Thru 
these points draw the usual 
measuring lines. Now, measur¬ 
ing from the line mn in the longi¬ 
tudinal section, take the various 
distances from the line to the 
points 5, 6, 4, 3, 2 and 1 in pro¬ 
file C, and place them upon similarly numbered measuring lines 
in pattern F, in each case measuring to the left from the stretch¬ 
out line m'ri', as shown. Connect these points, which will give 
the required miter cut upon the upper end of pattern for the 
common bar, laps being allowed, as shown by the dotted lines. 

The miter cut on the lower end of the common bar is obtained 
in precisely the same manner. Measuring from the line ho in 
the longitudinal section, take the various distances from this 
line to the points on the lower curb, profile B, and place these 
distances on similarly numbered measuring lines to the right of 
the vertical line h'o' in pattern F, thus obtaining the miter cut 



Problem 100. Flat Skylight. 





138 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


for the lower end of common bar A, where it intersects the lower 
curb B. 

The pattern for the miter cut on both ends of the lower 
curb will be the same, as shown in H, and is obtained by- 
placing the stretchout of profile B upon the line h"o" and 
drawing the usual parallel measuring lines, as shown. Then, 
measuring from the line ho in the sectional view, take the hori¬ 
zontal distances from this line 
to the points 10, 9, 8 and g in 
profile B, and place them upon 
similarly numbered measuring 
lines in pattern H, in each case 
measuring from the stretch¬ 
out line h"o". Connect these 
points, and as the lower curb 
from g to a will be a straight cut 
where it will butt against the 
upright member 7-6 in profile C, 
draw a vertical line from point 
g, completing the pattern, as 
shown. 

Problem 101. Double-Pitch Skylight 

Double-pitch, or gable, skylights (Figure 138) are used where 
a hipped skylight would be too expensive. In this form of sky¬ 
light, the ridge bar is extended to reach its full length, thus 
doing away with the end slopes and forming what is called a 
gable end. The gable ends are made of metal and, sometimes, 
of wood. Ventilators and louvres of any suitable form are 
frequently placed in the gable ends. The common bar is sup¬ 
ported at the top by the ridge bar and at the bottom on a curb, 
so designed as to receive the bar and to form a gutter to carry off 
the water that is emptied into it by the smaller gutters of the 


common bars. The pitch, or slant, of the skylight should never 
be more than a one-half pitch, altho a one-third pitch is well 
adapted to this form of skylight, and is the pitch in general 
use for both gable and hipped skylights. 

Figure 138 shows the half end elevation, profiles of the various 
parts, and patterns for a gable skylight having a one-third pitch. 
A sketch of plan view showing the position of the different 

members of the skylight and the 
curb dimensions, is shown in C. 
To construct the half end eleva¬ 
tion, first draw the center line 
6-e, and at any convenient point 
on this line locate the point g. 
From g at right angles to the 
center line 6-e, draw the line gc 
equal in length to one-half the 
width of the skylight, in this 
case 9 inches, thus locating the 
lower end of the glass line in 
common bar B at point c, as 
shown in lower curb F. 

The upper end of the glass line intersects the center line 6-e at 
point 5, which determines the pitch of the skylight, and is lo¬ 
cated in the following manner: (In this case, the skylight is to 
have the regulation pitch of one-third.) The line g-c is equal to 
9 inches, or one-half the width, or span. Double 9, which 
makes 18 inches, and divide by 3, the required pitch, which 
gives 6 inches. Now, measuring from point g, place this dis¬ 
tance on the center line, locating point 5, and draw the hypote¬ 
nuse of the right-angle triangle 5-g-c. Then, the line c-5 will 
represent one-third pitch. 

If, however, it was desired to make it one-half, or any other, 



Problem 101. Double-Pitch Skylight. 





SKYLIGHTS 


139 


Section Of Upper 
And Side Curbs 
6 


m 


Section Of Common Bar 





Section Of Lower Curb 



1 | 5 

/ 1 

"1 

1 

J 



J 

r 

!> 

i 

i 

! 

k 


' N 


3 

ry 

4 

N. \ 


5 r 1 

6 1 


6 

d‘ 


7 

i 

.j 

iZ 



\ 

Pattern For Upper 

And Side Curbs C 

3 




JO 


Fig. 137. Flat Skylight 


Upper Curb 


F 

Lower Curb 


n' 


o' 

h" 

7T 


H 


Pattern For Lower Curb B 


to 


/ 

7 / i 

/ 

b' 


/ 

2 

A 




u 

v\ 

3 


1 

I 

1 

J 

ry*~ 

4 

Measurinq Line. 



5 * 



‘ F 



5 

__ J 

c 

4 

Pattern ForCommonBar 

A 

1 

1 

J 

\ 

3 

/ 


< 

r 

f 


/ 




50 



















































































140 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


pitch, it is only necessary to divide the given span or width of 
the skylight by the number of the pitch to get the rise or height 
on the center line, as shown by g-5 in the elevation. For ex¬ 
ample, if the span of the skylight was 8 feet, a one-half pitch 
would rise on the center line one-half of 8, or 4 feet. Should a 
fourth pitch be required and the span is 8 feet, then one-fourth 
of 8 is 2 feet, or the rise from g to 5 on the center line. 

Next, draw the profile of the curb F, placing the edge of the 
glass line upon the point of the triangle at c, and taking the 
dimensions from profile F', make the formation of the curb, as 
shown. A tracing of one-half of the common bar B' is now 
placed in position upon the glass line c-5 of the end elevation, as 
indicated by the figures 1,2,3, h, 5 and 6 in profile!?, which is 
also a section of the gable. 

Take the various dimensions of the ridge bar from profile A' 
and construct the profile of this bar in its proper position, as 
shown at A. When constructing the profiles at A and F, the 
student must be careful to see that the rest for the glass is in the 
same plane as the rest for the common bar B. 

The pattern for the gable end can be made in one piece, but it 
is best to make it in two pieces, as it can be cut from the metal 
with less waste and formed much more easily upon the brake. 
A one-half pattern for the gable end is shown in G, and'is ob¬ 
tained as follows: First, draw the stretchout line SV at right 
angles to the glass line in the elevation, and upon this line place 
the stretchout of the half bar, as shown from 1 to 6 in profile!?. 
Thru these points, draw the usual measuring lines, which inter¬ 
sect by lines drawn from similarly numbered points of intersec¬ 
tion of the half bar with the ridge and curb, as shown. Next, 
from point 6' of the pattern, draw the vertical center line 6'-x, 
and upon this line locate the point 7', making the distance 6'-7 f 
equal to the distance 6-7 on the center line in the end elevation. 


Now, at any convenient distance from the center lines in both 
pattern G and the end elevation, draw the vertical lines mn; in 
each case, they must be the same distance from the center line. 
The stretchout of the lower part of curb F is now placed upon 
the line mn in pattern G and measuring lines drawn, as shown 
from 7 to 12. 

Measuring in each case from the line mn in both elevation and 
pattern, take the various distances to the points on curb profile 
F and place them upon similarly numbered measuring lines in 
pattern G. Connect these points, completing the half pattern 
for gable. 

The pattern for the miter cut on curb F is shown in E. The 
stretchout of profile! 1 is placed upon the line mn, and the miter 
cut from 7 to 12 is obtained in precisely the same manner as the 
lower part of pattern G, the upper part of the pattern from 7 to a 
being simply a straight cut. 

The pattern for the common bar is shown in H, and is ob¬ 
tained as follows: At right angles to the line 6-6" in the end 
elevation, draw the line JL, upon which place the stretchout of 
the common bar, as shown by the figures 6 to 1 in profile!?, and 
similar figures on the line JL. Thru these points, and at right 
angles to JL, draw measuring lines and intersect them by lines 
drawn at right angles to 6-6" from the points of intersection on 
the ridge bar and curb, as shown. A line traced thru these 
points will be the pattern for the common bar. 

Problem 102. Hipped Skylight 

A hipped skylight is one that has an equal slope or pitch on all 
of its sides. The curb forming a continuous molding that passes 
horizontally around the four sides of the base is usually set on a 
level roof curb, for the four glazed sides provide the necessary 
pitch to shed the snow or rain. 


Side Curb 


SKYLIGHTS 


141 









































































142 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Figures 139 and 140 show the method for obtaining the pat¬ 
terns for a rectangular or square hipped skylight having a one- 
third pitch. The same principles are applicable to all hipped 
skylights, no matter what the pitch of the skylight may be or 
what angle its base may have. The drawings are one-fourth 
full size. When the full-size patterns are developed, they will 
be of such size as can 
be used for practical 
work. 

In Figure 139 is shown 
a one-half sectional 
view, a quarter plan, 
and patterns for the 
curb, ridge, jack and 
common bars; also a 
diagram showing a de¬ 
tailed plan of intersec¬ 
tions of the various bars 
used in the construction 
of hipped skylights. 

First, draw any cen¬ 
ter line, as 9-x, at right 
angles to which draw 
the horizontal line A-2 equal to 10 inches, or one-half the 
width or span of the skylight. Double 10, which makes 20, and 
divide by 3 or the pitch required, and place this distance, 
which is the height or rise, upon the vertical center line from A 
to 2', as shown. Now, draw the hypotenuse of the right-angle 
triangle 2'-2, which represents one-third pitch. At right angles 
to 2'-2, place a section of the common bar, as shown by A; thru 
this draw lines parallel to 2'-2, intersecting the curb B at the 
bottom and ridge bar F at the top. 


Next, draw the section of curb!?, placing the edge of the glass 
line upon the point of the triangle at 2 and the curb line gh di¬ 
rectly below it, making the formation of the curb, as shown. A 
section of the ridge bar is next placed in position, as shown from 
9 to 8 in F. Number the corners of the common bar A, as 
shown, from 1 to 6 on each side, thru which draw lines parallel to 

2'-2 until they intersect 
the curb at the bottom 
and the ridge bar at the 
top, as shown by similar 
figures 1 to 6. The 
member 5-6, or gutter 
side of the bar, is cut 
away at any convenient 
distance from the end 
to permit of the free 
discharge of the water 
from the common bar 
into the gutter of the 
curb. This completes 
the one-half sectional 
view of the skylight, 
from which the pattern 
for the common bar can be developed as follows: At right 
angles to the line 2'-2, draw the line mn, upon which place 
the stretchout of profile A, as shown by the figures 1 to 6. 
Thru these points draw the usual measuring lines, and intersect 
them by lines drawn at right angles to 2'-2 from similarly num¬ 
bered intersections on curb B and ridge bar F. Then a line 
traced thru the various points of intersection will be the pattern 
for the common bar. 

The pattern for curb B is obtained by taking the stretchout of 



Problem 102. Hipped Skylight. 




SKYLIGHTS 


143 




















































































144 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


the various corners in the curb, a, b, 2, 3, k, C, e, g, h and o, and 
placing them on the vertical line m"n" as indicated by similar 
letters and figures in the pattern. Thru these points draw hori¬ 
zontal measuring lines, as shown. Next, thru point C in the 
curb profile B, draw the vertical line mV. Measuring from the 
line mV, take the horizontal distance to the various points on 
curb profile B and place them on similarly numbered measuring 
lines in the pattern, measuring in each case from the stretchout 
line m"n", as shown. A line traced thru these points will be the 
half pattern for the curb A, condensation hole to be punched 
into the curb between each light of glass is shown in position 
above line U in the pattern. 

Jack Bar .—Before the patterns for the hip and jack bars can be 
developed, a quarter plan view must be drawn, which will give 
the points of intersection between the hip and jack bar, between 
the hip and ridge bar, and between the hip and curb. As the 
skylight forms a right angle in plan, from any point as S on the 
center line 9-x, draw a line at an angle of 45° and intersect it by 
a vertical line drawn from point 1 in curb B, at 1 in plan. Then 
the diagonal line S-l represents the hip line in plan. 

A profile of the common bar A is now placed on the hip line 
S-l so as to obtain the horizontal measurements, being careful 
to have the points 1, 2, 4 come directly on the hip line S-l. 
Thru the various points on the profile, lines are drawn parallel to 
the hip line S-l , which intersect vertical lines drawn from simi¬ 
larly numbered points in curb# and ridge barF in the sectional 
view. A line traced thru these points, as shown from 1 to 6 in 
plan, will represent the intersections between the hip bar and 
curb and ridge bar. 

Before the pattern for the jack bar can be developed, the miter 
line between the jack bar and hip bar must be obtained, both in 
the elevation and plan, in the following manner: Take a 


tracing of the common bar, profile A, and place it in a horizontal 
position in plan, as shown. Thru the various points in the pro¬ 
file draw horizontal lines, intersecting similarly numbered lines 
on one side of the hip bar in plan, thus forming the miter line of 
the short cut from 1 to 6; also the miter line of the long cut from 
1 to 6'. From these intersections at right angles to the lines of 
the jack bar, draw vertical lines intersecting similarly numbered 
lines in the half sectional view, obtaining the points of intersec¬ 
tion and miter line of the short cut, shown by 1,2, 3, U, 5 and 6, 
and the long cut, shown by 1,2,3', U, 5' and 6'. 

For the pattern for the upper cut of the jack bar, the same 
stretchout can be used as that used for the common bar. There¬ 
fore, from the various intersections on the miter line in the sec¬ 
tional view and at right angles to the glass line 2'-2, draw lines 
intersecting similarly numbered lines in the pattern for the com¬ 
mon bar. A line traced thru these points of intersection will be 
the pattern for the upper cut of the jack bar, the lower cut being 
the same as that shown on the pattern for the common bar. 
Ridge Bar .—The pattern for the ridge bar is simply a piece of 
metal, rectangular in form, its width being equal to the stretch¬ 
out of profile F, as shown from 8 to 8 on the line mn in the pat¬ 
tern. The length of the ridge bar is equal to the difference 
between the length and width of the curb of the skylight. 

Hip Bar .—Before the profile and pattern for the hip bar can 
be developed, a true elevation of the hip bar must be con¬ 
structed, as shown in Figure 140, the one-half sectional view 
and quarter plan being a duplicate of Figure 139. The true 
elevation of the hip bar could be drawn in Figure 139, but in 
this case, for want of space and to save overlapping of lines, it 
has been transferred to Figure 140, as shown. 

The true intersections between the hip bar and curb and ridge 
bar being shown in their proper position in the quarter plan, the 


SKYLIGHTS 


145 



































146 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


next step is to draw a true elevation of the hip bar, from which a 
true profile of the hip bar and pattern are obtained as follows: 
Parallel to the hip line S-l in plan, draw any line, as CG. From 
the various intersections 1 to 6 in both curb and ridge in plan, 
draw lines at right angles to S-l , crossing the line CG indefinitely. 
Since the several points in the upper miter of the hip bar must 
appear in the true elevation of the bar at the same heights as 
corresponding points of the common bar in the one-half sectional 
view, these distances may be transferred from the sectional 
view to the true elevation of the hip bar by means of the dividers. 
Now, measuring in each and every instance from the line A-2 in 
the one-half sectional view, take the vertical heights to points 1 
to 6 in the ridge and 1 to 6 in the curb, and. place them in the 
elevation of the hip bar on similarly numbered lines drawn from 
the plan, measuring in each instance from the line CG. Thru 
these points, draw the miter lines and connect similarly num¬ 
bered points at the top and bottom by lines, as shown. If the 
miter lines at the curb and ridge in the elevation are true, these 
lines will all run parallel to the glass line 2-2, as shown, which 
completes the true elevation of the hip bar, intersecting the ridge 
and curb. 

It is next necessary to determine the profile of the hip bar, 
shown at R. This is accomplished by drawing the line OV 
parallel to the glass line 2'-2 in the one-half sectional view. 
From the various points on profile A and at right angles to OV, 
draw lines to intersect the line OV, which will show the widths of 
the various parts of the bar, as shown by the figures 1 to 6 on the 
line OV. Now, take the various divisions on this line and place 
them upon the line O'V', which is drawn parallel to the lines of 
the hip bar in the true elevation. From the various points on 
this line at right angles to the line 2-2, draw lines intersecting 
similarly numbered lines in the elevation of the hip bar, as shown 


from 1 to 6. Connect these points on either side, as shown; 
then R will be the true profile of the hip bar. 

Pattern for HipBar .—The pattern for the hip bar is developed by 
taking the stretchout of profile R and placing it on the line ab 
drawn at right angles to the lines of the hip bar. Thru the vari¬ 
ous divisions on the stretchout line ab, draw the usual measuring 
lines, which intersect by lines drawn at right angles to the line 
2-2 in the elevation from similarly numbered points in the miter 
lines of the ridge at the top and the curb at the bottom. A line 
traced thru these points will be the pattern for the hip bar. 

Bar Intersections .—The intersections of the various bars with 
the ridge as they would appear in the plan view of a hipped 
skylight, are shown in diagram R, Figure 139. The ridge bar G 
is intersected by the common bar F, the miter cut being shown 
by ebe in the pattern for the common bar. The jack bar E 
intersects one side of hip bar A, shown by the short miter line 
no and the long miter og. The long and short miter cuts are 
shown by similar letters in the pattern for the jack bar. 

If the centers of the common jack bars C are mitered, as 
shown at b, one side of the bar will intersect the ridge, as shown 
by the miter line be, while the other side will intersect the hip bar 
A, shown by the miter line bs; then the miter cut eb would be 
the same as the cut on one-half of the common bar, and the 
miter cut bs the same as the cut on the long side of the jack bar. 
Then, the pattern for common jack bar C is simply a common 
bar pattern having the cut for the long side of a jack bar placed 
upon one-half of the pattern, as shown by the dotted lines bs' on 
the upper end of pattern for the common bar. 

The two hip bars A and A in diagram R intersect the ridge 
bar, as shown by the dotted miter line ab, the inner half of the 
bars intersecting on the line bt. The miter cuts tb and ba are 
shown by similar letters on the upper end of the pattern for the 


SKYLIGHTS 


147 


hip bar. The center jack bar B intersects the two hip bars A 
and A on the miter lines bm, and the pattern is simply a jack 
bar pattern having a long cut on each side of the center 
line, as shown by og and Og' in the pattern for the jack bar. 

Square hipped 
skylights are often 
required in practi¬ 
cal work, and the 
intersections of the 
four hip bars are 
shown by the miter 
lines ac in diagram 
A, Figure 140. The 
miter cuts ac on 
each side of the in¬ 
tersected hip bars 
are shown by tbm on 
the upper end of 
the pattern for the 
hip bar. 

Diagram B shows 
the intersection of 
hip bars F and F 
with a ridge venti¬ 
lator, the profile of 
the lower part of 
the ventilator being the same as one-half of ridge bar F. The 
hip bar pattern with the miter cut ba placed upon each side 
of the center line, will be the pattern for the hip bar F in 
diagram B. 

Measurements .—Having now developed the patterns for the 
various parts of the hipped skylight, it is usual in actual shop 



Fig. 141. Rule for Finding Factors 


practice to transfer the various patterns from the drawing to 
sheet metal by means of a prick punch, marking the measuring 
points and the name of the pattern upon each one. The metal 
patterns are kept for future use and can be used when con¬ 
structing any style hipped skylight having a one-third pitch, no 
matter what the size of the curb may be. 

The size of the curb, also the widths of the ventilators, when 
used, forms the basis for obtaining the lengths of the various 
parts of a hipped skylight. There are several methods used for 
finding the true lengths of the ventilator, common, jack, and hip 
bars; but the best results are obtained by mensuration, in which 
a factor or multiplier is used to find the length of the bars with¬ 
out using any drawings, diagrams, scales, triangles, etc., when 
the size of the curb and width of the ventilator are known. 

The following rules should be observed for finding the true 
lengths of the various parts of a hipped skylight by mensuration: 

1) To find the true length of the ridge bar, deduct the shortest 
side of the curb from the longest side. 

2) To find the true length of the ventilator, deduct the shortest 
side of the curb from the longest side and add the width of the 
ventilator. 

3) To find the length of the common bars for a one-third pitch 
skylight, divide the short side of the curb by 2 and multiply by 
the factor 1.2. 

4) The distance on the curb between the hip and jack bars, 
multiplied by the factor 1.2, will give the true length of jack bars 
for a one-third pitch skylight. 

5) To find the true length of the hip bars for a skylight having 
a one-third pitch, divide the short side of the curb by 2 and mul¬ 
tiply by factor 1.56. 

6) To find the true lengths of the common, hip and jack bars 
in a hipped ventilating ridge skylight, deduct the width of the 















148 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


ventilator from the short side of the curb, then divide by 2 and 
multiply by the same factors used for one-third pitch sky¬ 
lights. 

The factors here given can be used only for skylights having 
one-third pitch; also the curb line must always run in line with 
the edge of the glass line, as shown by 2, g, h, in Figure 140. 

The rule for finding the factors used for one-third pitch is 
shown in Figure 141. If the rise of the common bar, shown by 
am, is 8 inches, and the base mn is 12 inches, the length of the 
hypotenuse an will equal the square root of the sum of the 
squares of the rise and base, or 14.4222 inches. Now, divide 
this length 14.4222 by 12, the length of the base, which gives 
1.2018. This number and decimal will be the factor to use for 


finding the length of the common and jack bars. The factor for 
the hip bar is found in a similar manner. As the length of the 
hip bar, eg in plan, equals the square root of the sum of the 
squares of the two sides ec and eg, the true length of the hip bar 
shown by a'g' in the diagonal elevation, will equal the square 
root of the sum of the squares of the two 12-inch sides and the 
square of the 8-inch rise, shown by e'a', thus: V 12 2 + 12 2 = 
V 288 + 64 = V 352 = 18.7616 inches. 

Now, divide 18.7616 by 12 and the quotient 1.5634 will be the 
factor to use in finding the length of the hip bars. In practice, 
1.56 is used. Should the skylight have a different pitch or rise 
from 8 inches to 1 foot of base, the factors can be found in a 
similar manner. 





CHAPTER X 


Special Problems 


Problem 103. Rectangular Flaring Pan with Folded 
Corners 

The sheet-metal worker is often required to construct square 
and rectangular flaring pans, the corners of which are to be made 
water-tight by folding together and turning them to the sides or 
ends of the pan, having the upper edges of 
the folded parts finish neatly under the 
wire which is enclosed along the top. 

A pan constructed in this manner is 
water-tight without soldering, and, since 
this style of pan is often subjected to a 
temperature higher than the melting point 
of solder, the construction shown in Fig¬ 
ure 142 is important. In actual shop prac¬ 
tice, the pattern is laid out with a steel 
square and the dividers directly upon the 
sheet of metal. 

In Figure 142 is shown the full pattern 
and side elevation of a rectangular flaring pan having an equal 
flare on all sides. Draw the side elevation, shown by ACBG, 
the lines AB and CG show the slant height of the pan. Next, 
draw the plan of bottom E HDF. From each comer of the plan 
extend the lines, as Hn, En, etc., in each case making the dis¬ 
tance from the corner to n equal to the slant height of the 
sides, shown by GC in the side elevation. Now, thru these 



Problem 103. Rectangular Flaring Pan 
with Folded Corners. 


points parallel to EH and EF, draw lines as am, which intersect 
by vertical lines drawn from A and C in the side elevation. Con¬ 
nect the corners E to a and H to m. Since the flare of all 
sides of the pan are the same, as shown by an and nm, place 
these distances upon the line nm at each of the four corners 
and draw the miter lines, as shown. Then 
will amstxdfb represent the pattern for the 
pan if the corners were butted together. 
The allowance for wire is now added, as 
shown by the dotted lines. 

The next step is to provide for the extra 
material required for a folded corner, and 
the manner in which this result is accom¬ 
plished is shown in the upper left-hand 
corner of the pattern. Bisect the angle 
Eab, obtaining the line of bisection RE. 
Set the compasses to a radius slightly less 
than the distance from the point a to the 
line RE, and from a as center, describe the short arc oc, 
intersecting the flare line aE at g. With g as center and gc as 
radius, intersect the arc oc at e. Draw a line from a thru e, 
producing the line until it meets RE in the point h. From h 
draw a line to b. Then Rbha is the amount to be notched 
from the corners, so that when folded, they will finish neatly 
under the wired edge of the pan. 


149 




150 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 



Problem 104. Round Ventilator with Molded Base 

This problem is presented as an example of the practical appli¬ 
cation of the three methods of develop¬ 
ment employed in developing patterns 
for previous problems in this course. 

Since the drawings represented are those 
commonly used in a sheet-metal shop, 
the student is expected to construct the 
various patterns, aided only by the in¬ 
structions he has already received. 

Figure 143 shows the elevation of a 
cupola and ventilator with a square 
molded base that fits over a double- 
pitched roof. The drawing is shown 
one-fourth full size, and the different 
parts of the article are indicated by 
letters, as shown. The lower portion of 
the ventilator, shown in F, is square in 
form and fits over a roof having a one- 
third pitch. A tapering transition piece 
from square to round, shown in A, is 
required between the square base and 
the round pipe, shown in section G. Sec¬ 
tion H is the frustum of a cone whose 
sides flare at an angle of 45° with the 
horizontal, and which is placed over 
the upper edge of the round pipe, as 
shown. The angle of inclination of 
upper section C is 47^ ° to the horizon¬ 
tal, and is joined at the top by the round tapering piece B and 
the ball E. A circular band or windshield is shown by ache 
in the elevation. The band is strengthened by enclosing a wire 


in its upper and lower edges, and is fastened to the upper 
flange C and lower flange H by means of a band iron brace, as 
shown in the drawing. Draw the full 
size elevation, and, after the patterns 
have been developed for the various 
sections, the customary allowances for 
laps or riveting edges should be added 
to the drawing. 

Problem 105. Ball, Development 
by Zones 

This problem is illustrated by Figure 
144. The sphere is the most prominent 
example of the many forms whose sur¬ 
faces admit of approximate develop¬ 
ment. Since solids of this kind may be 
resolved into parts resembling those 
capable of accurate development, the 
same general methods are applicable. 
Thus, in the case of the ball, shown in 
Figure 144, this solid has been resolved 
into a number of frustums of cones, 
and patterns developed in the regular 
way. This method is called develop¬ 
ment by zones or horizontal sections, 
and may be used for obtaining patterns 
or blanks for any curved work, whether 
hammered by hand or machine. 

The ball can also be made from gores 
or vertical sections, but many sheet-metal workers prefer to 
make them from zones, as the pieces are not so apt to warp, 
and the labor is less in raising the zones than in raising the gores. 


Problem 104. Round Ventilator with Molded Base. 








SPECIAL PROBLEMS 



Fig. 142. Rectangular Flaring Pan 


151 

































































152 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


In Figure 144 is shown the elevation and patterns for a ball 
constructed by means of zones. First, draw the elevation of the 
ball the required size and divide into as many zones as desired- 
in this case, six. When the ball is large, more zones must be 
used so as to require the least amount of labor in raising the 
blanks to form, for 
it takes less time 
to cut an extra 
blank and raise the 
ball to a truer cir¬ 
cle than to use less 
zones and raise 
them to a greater 
depth. 

Having divided 
the quarter circle 
into three equal 
parts, shown by 
abeg, the radii with 
which to develop 
the patterns for 
zones A, B and C 
are obtained as fol¬ 
lows: Thru the 
center of the ball, 
draw the vertical line F H; then, thru the extreme points of 
the zones A and B, draw the lines ge and eb, which extend until 
they meet the center line at F and G, respectively. Where the 
two zones A and B joint together on the line el , use m as a center 
and describe the quarter circle 1-5-m, which represents the half 
section thru 1-m. Divide the half section into a number of equal 
spaces, as shown from 1 to 5. With F as center and radii equal 


to Fe and Fg, describe the arcs e'f and g'h. Now, draw any 
radial line, as g'f, and, starting from e' on the upper arc, step off 
twice the number of spaces contained in the section 1-m, as 
shown from 1 to 5 to 1 in the pattern. From F thru /, draw a 
line intersecting the arc g'h at h, completing the half pattern for 
zone A, to which laps are allowed, as shown. To obtain the 
pattern for zone#, use n as center, and with radii equal to Gb and 
Ge in the elevation, describe the arcs b' and cc'. Draw the ra¬ 
dial line cn, and, starting from c, step off on the center arc four 
times the number of spaces contained in the section 1-m, and 
draw a line from c' to n, intersecting the inner arc at b'. Then 
cb''c'b' is the full pattern for zone B. The upper zone C is 
formed of a circular disc of metal; the pattern is described with a 
radius equal to ab in the elevation. When constructing the 
drawings, no attention is paid to the lower half of the ball, since 
the patterns for the upper half will serve, also, for the lower half. 
After the patterns have been formed to a true circle and soldered 
together on the inside, they are subjected to the raising process, 
and the desired curve is made with the raising hammer upon a 
wood or lead block. 

Problem 106. Ball. Development by Gores 

In Figure 145 is shown the approximate method for develop¬ 
ing the patterns for a ball composed of eight vertical sections, as 
shown in the plan view. This method is referred to as the de¬ 
velopment by gores. The patterns are obtained by means of 
parallel lines as follows: 

Let B represent the plan view of the ball, which is divided into 
as many parts as gore pieces required. In this case, the ball is 
to have eight gore pieces, altho any number of pieces can be 
used, and the principles will apply. With 1 '-m' as radius and m 
as center, describe the quarter circle m-1-7, which represents 



Problem 105. Ball, Development by Zones. 



SPECIAL PROBLEMS 


153 











































154 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


the half section thru l'-m', shown in A. Divide the half sec¬ 
tion into a number of equal spaces, as shown from 1 to 7, 
and from these points draw horizontal 
lines, intersecting the miter lines l'-a 
and l’-b in plan, as shown. At right an¬ 
gles to ab draw the center line 1 '-c, upon 
which place twice the number of spaces 
contained in half section A, as shown from 1 
to 7 to 1 in pattern G. Thru these points 
draw lines at right angles to l’-c, which 
intersect by vertical lines drawn from simi¬ 
larly numbered points on the miter lines 
l’-a and l’-b. A line traced thru these 
intersections, as shown in G, will be the 
desired pattern. A small edge should be 
added to one side of the pattern, so that the 
gores can be joined together by means of a 
lapped seam. 

Problem 107. Base for Chimney Top, 

Rectangular to Round 

In Figure 146 is shown a short method 
for laying out the pattern for the base of a 
chimney top in the form of a transition 
piece from rectangular to round, which is 
to be constructed in two pieces. In actual 
shop practice, the pattern is laid out with 
the steel square directly upon the sheet of 
metal. Let ABCD represent the portion of 
a sheet of iron, which can be of any desired width. The rec¬ 
tangular base in this case is assumed to be 13 inches by 17 
inches, and the diameter of the round top 7 inches. About 5 


inches from the bottom of the sheet, draw the line mn parallel 
to BD, and upon this line locate point a, making the distance 
via a little more than one-half of the width 
of the short side of the base. Next, measur¬ 
ing from a, make ab equal to the length of 
the long side of the base; in this case, 17 
inches. Bisect ab and erect the perpen¬ 
dicular, as shown by the center line ge. 
Now, measuring from point e on the upper 
edge of the sheet of iron, locate the points 
o and h, making the distance oh equal to 
one-half the circumference of the top, after 
which the outer edge of the steel square is 
placed upon the points o and a, being care¬ 
ful to have the 6-1/2-inch mark on the short 
arm cf the square directly over point a. 
Lines are then scribed upon the metal, as 
shown by ot and ta. Now, move the steel 
square along the line ot to x, making the 
distance tx equal to m B, which is the width 
of the vertical flange that extends down 
the side of the chimney. Then draw the 
lines txv. After this, reverse the square 
to the position, shown in S, to obtain the 
cut for the corners b and a. The opposite 
side of the pattern is treated in the same 
manner, and laps are added, as shown by 
the dotted lines. The circular cut on the 
upper edge of the pattern is obtained by 
extending the line xo and x'h until they intersect at F. With F 
as center, and Fo as radius, describe an arc, which will be the 
upper edge of the pattern. This completes the half pattern for a 



Problems 107 and 108. Base for Chimney 
Top and Hood. 













LUD 9g doj_ 


SPECIAL PROBLEMS 


155 


u/vsQ apt g 



1 

X 


c 

e 

•** 

JS 

CJ 

^1- 


64 


£ 




< 


Fig. 146. Base for Chimney Top 
































156 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


chimney base, rectangular to round, made in two pieces; the 
other piece can be cut from the same sheet of metal without any 
waste by simply reversing the position of this pattern when 
placing it upon the sheet. 

Problem 108. Chimney Hood 

Figure 147 shows the pattern for a one-piece chimney cap 
which can be laid out directly upon 
the metal. First, lay out the pat¬ 
tern for the round pipe of the width 
and circumference required, as shown 
by ABCD. Bisect the line AC and 
erect the perpendicular mn. With 
m as center and mA as radius, describe 
the arc An. Thru point n draw a 
horizontal line parallel to AB, which 
intersect by the vertical lines BA and 
DC extended to F and G. Next, divide 
the two spaces Am and Cm into three 
equal parts, as shown. With the 
dividers set equal to one of these 
spaces, as m-2, and using points CA 
and m as centers, describe the half 
circle 2-a-3 from m, and the quarter 
circles b-1 and ^-e from A and C, as 
shown. Cut down the center line from n to m: then cut out 
the half circle 2-a-3 and the quarter circles A-b-1 and C-J^-e. 
Laps are allowed for seaming on the top and sides, as shown. 
Form up the pipe in the forming rolls, and groove the edges 
AB and CD. Then form the top and make a grooved or riveted 
seam on the edges Gn and nF. This cap can be quickly made 
without waste of material, and is frequently used as a cure for 


chimneys that are troubled from the rebound of the wind 
against a higher building. When used for this purpose, the 
side of the cap is set parallel with the side of the building. 

Problem 109. Florist’s Watering Pot 

In Figure 148 is shown the elevation and patterns for the va¬ 
rious parts of a florist’s watering pot. The body of the sprinkler, 
shown at A, has a wire enclosed in the 
top at mn, the lower edge being con¬ 
nected to the bottom fv by means of 
a double seam. At F is shown the 
tapering spout which joins the body at 
s and t. The sprinkler rose R is 
soldered to the tapering tube or socket 
H, which slips down over the end of 
the spout F to hold the sprinkler rose 
in position, and also allows it to be 
removed for cleaning and other pur¬ 
poses. The circular handle B with its 
hand grasp or boss on the inner side, 
shown at h, is placed over the center 
of the top; the ends being attached to 
the body in the front and rear, as 
shown at m and g. 

The helmet lip or breast C has a 
wired edge, shown by db, the lower edge mb is soldered to the 
wired edge of the body on the line mn, as shown. This is the 
most common form of breast which is added to certain vessels 
for liquids, particularly such as are used for pouring. 

The pattern for the breast is shown in C' and is developed by 
the parallel-line method as follows: With b as center and bm as 
radius, describe the quarter circle me, which will represent a one- 



Problera 109. Florist’s Watering Pot. 




SPECIAL PROBLEMS 


157 



. 





























































158 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


half plan of the breast. Divide the quarter circle me into a con¬ 
venient number of equal spaces, as at 1, 2,3, A, 5 and 6. From 
these points erect vertical lines, which intersect the base line 6m, 
as shown. Now, from the points on the base line mb, parallel to 
the line md, draw lines indefinitely intersecting the vertical line 
db, as shown. Draw the perpendicular line a'6'; then, measur¬ 
ing from the line mb, 
take the various dis¬ 
tances to the points 1, 

2,3, etc., on the quar¬ 
ter circle me, and place 
them on similarly 
numbered parallel 
lines extended in sec¬ 
tion G, as shown. A 
line traced thru these 
points will complete 
the full view of a cross 
section of the breast, 
as it would appear if 
taken on the line db in 
the elevation. The 
pattern is now ready 
for development by 
the parallel method. 

From point 6 at right angles to md, draw the stretchout line b-6, 
upon which place twice the number of spaces contained in section 
G, as shown from 6 to 1 to 6. Thru these points, draw the usual 
measuring lines, which are intersected by like numbered lines 
drawn at a right angle to the line md from similarly numbered 
points on the lines mb and bd. A line traced thru these inter¬ 
sections will be the pattern for the breast, as shown in C. 


The outline of the tapering spout is shown in F. Complete 
the full cone in the elevation, as shown by the dotted lines, and 
produce them to the apex at x. Next, draw the plan and de¬ 
velop the pattern for the spout, shown at F, by the method de¬ 
scribed in Problem 63. 

The sprinkler rose R and the tapering socket H are the frus¬ 
tums of two cones, and the patterns are developed by the radial 
method described in Problem 43. The pattern for the circular 
handle# is shown at#', and is a strip of metal equal in length to 
mBg in the elevation. 

At E is shown the method of laying out the pattern for the boss 
on the inner side of the handle, shown at h in the elevation. This 
piece is added so that the handle may more readily be grasped 
and held firmly in the hand. Take a tracing of the boss and 
place it in position on the arc m"g, as shown. The curved out¬ 
line, shown in K, is a section thru the central portion of the 
handle. Divide the section into equal spaces, as shown from 1 
to -4- Thru these points, draw vertical lines, intersecting the 
outline of the handle on the arc m'g', as shown. The girth of 
section K is then placed upon the stretchout line, and the pat¬ 
tern is obtained by the method of parallel developments, shown 
at E. 

Problem 110. Automobile Measure 

Figure 149 shows the method of developing the patterns for a 
liquid measure or pouring-can having an offset funnel attached 
to the top. First, draw the elevation of the flaring measure with 
the funnel attached, in accordance with the dimensions given in 
the drawing. The patterns for the body of the measure and the 
spout for the funnel, shown by a-b-1 -7, are developed by the 
radial method and no further explanation is necessary. The 
pattern for the spout is shown at R. 

The pattern for the funnel A, shown by 1,7, 8,12', 16', is de- 



Problem 110. Automobile Measure. 








SPECIAL PROBLEMS 


159 


























































160 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


veloped by the short method of triangulation described in 
Chapter VII. Construct half sections on the funnel, as shown 
by CB and F. These half sections represent the shape of the 
funnel on their respective lines. Divide each section into equal 
spaces, and from these points draw perpendicular lines, which 
intersect the base 
lines 1-7, 8-12', 
and 16'-12', as 
shown. Then 
connect the vari¬ 
ous points on the 
base lines with 
the usual solid 
and dotted lines, 
shown in A. The 
true lengths of 
the solid and 
dotted lines are 
shown in dia¬ 
grams H and D, 
and are found in 
the customary 

manner, as de Problem 111. A Pieced Elbow Intersecting 
scribed in Prob- a Round Pipe, 

lem 80. 

The full pattern for the funnel is shown in J, and is developed 
in precisely the same manner as described in connection with 
Figure 124. The seam line 7-8 and the center line 1-16 in the 
pattern are equal to 7-8 and 1-16' in the elevation. The divisions 
from 1 to 7 in the pattern are equal to the divisions from 1 to 7 
on half-section C. The divisions 8 to 12 on the lower edge of the 
pattern are equal to the divisions 8 to 12 in half-section B. The 


divisions from 12 to 16 on the upper edge are equal to the spaces 
from 12 to 16' on the outline of half-section F. After the pat¬ 
tern is constructed, edges are added for seaming and wiring. 

Problem 111. A Pieced Elbow Intersecting a 
Round Pipe 

In the sheet-metal worker’s experience, it is often necessary to 
connect two pipes of different diameters whose axes lie at right 
angles to each other. When making a connection of this kind 
in blow-pipe work, it is important that the fitting be so con¬ 
structed as to secure an easy flow of air thru the pipes. A fitting 
that will accomplish this result is shown in Figure 150. 

The vertical pipe in this case is assumed to be 6 inches in 
diameter. It is connected to a horizontal pipe 4-3/4 inches in 
diameter by means of a four-piece elbow, shown by K HFB. 
First, draw the angle agb and make it equal to that of the re¬ 
quired fitting; that is, 90°. Next, on the line gb lay off a dis¬ 
tance of 5 inches from g to e, which is the radius desired for the 
inner curve of the elbow. Make eb equal 4-3/4 inches, the diam¬ 
eter of the elbow, and, with g as center and ge as radius, describe 
an arc and construct the full elevation of the elbow, as shown. 
Since this process has been fully described in Problem 16, the in¬ 
structions need not be repeated here. After the elevation of the 
elbow has been drawn, an elevation of the vertical pipe is placed 
in position, as shown by ABCD. Draw the profile of the elbow, 
as shown atG, dividing it into equal parts, as shown from 1 to 7. 
Draw a plan view of the vertical pipe, shown at R, and place 
duplicate of profile G in elevation in its proper position, as shown 
by G' in plan. From the points 1 to 7 in profile G', draw hori¬ 
zontal lines to the right until they intersect the outline of the 
larger pipe, from which intersections erect vertical lines indef¬ 
initely. . From the various points in profile G in elevation, draw 

























































162 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


horizontal lines until they intersect the miter lin efh, afterwards 
producing them in the manner shown in H and F until they in¬ 
tersect corresponding vertical lines previously drawn from the 
plan. A line traced thru these intersections 
will represent the miter line between the ver¬ 
tical pipe and the sections H and F of the 
elbow. This irregular curve crosses the 
miter line st at m, and this point is projected 
to profile G in the elevation and profile R in 
the plan, as shown at m and m'. These 
points will be used in developing the pat¬ 
terns. It will be seen if the drawings are 
carefully studied at this point, that the por¬ 
tion of the elbow shown by the dotted lines 
is not required. Patterns for a small portion 
of section F, a certain amount of the surface 
of section H, and the entire surface of sec¬ 
tion K, will be required. The pattern for 
section H is shown in E and is developed in 
the following manner: Draw any vertical 
line, as a'b' in the pattern, upon which place 
the stretchout of profile G in the elevation. 

Thru these points, at right angles to a'b', 
draw measuring lines, as shown. Next, from 
point / in section H in the elevation, draw 
the line fd at right angles to hs. Now, meas¬ 
uring from the line fd, take the various dis¬ 
tances to the points on the miter lines hf and t-m~b, and place 
them on similarly numbered measuring lines in pattern E, 
measuring in each case on either side of the line a'b', as 
shown. A line traced thru these points will complete the 
required pattern. The curved outline of pattern E, shown 


by lF-h'-lF, will also serve as the miter cut for section K. 

When developing the pattern for section F, shown at P, it will 
be seen that it only requires the stretchout of profile G from 
point m to IF to m', and that point k! repre¬ 
sents the position of the central and longest 
measuring line of the pattern. Place the 
stretchout upon the line m'm, and develop 
the pattern in -the manner described for 
pattern E, measuring from the line tn in 
section F to the various points on the 
miter line mt and the curved line of inter¬ 
section m-l+. 

The pattern for the opening in the straight 
pipe is laid off at T; the divisions from 1 to 7 
on the horizontal stretchout line are equal 
to similarly numbered divisions on the out¬ 
line of the larger pipe in plan. From the 
points on the stretchout line, erect vertical 
lines, which are intersected by horizontal 
lines drawn from similarly numbered points 
on the curved miter line in the elevation, 
resulting in the pattern shape, shown in T. 

Problem 112. Round Ventilator with 
Square Base 

This problem is presented to show the 
method employed in developing the patterns 
for a quadrangular pyramid intersected by a round pipe. 

In Figure 151 is shown the elevation and plan view of a round 
ventilator with a base in the form of a square pyramid that fits 
over a double-pitched roof. The drawing is shown one-third 
full size. The various parts of the ventilator are indicated by 



Problem 112. Round Ventilator with 
Square Base. 










SPECIAL PROBLEMS 


163 




















































164 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


the letters, as shown. The upper portion of the ventilator is 
round in form and consists of the flat cone top, shown in A ; the 
flaring flange or fustrum of a cone, shown in B, which is con¬ 
nected to the upper edge of the round pipe F. A circular band 
or windshield, shown by ageb, is connected to the lower flange 
B and the top section A by means of a band iron brace, shown by 
dotted lines in the elevation. 

The patterns for sections A and B are developed by the radial 
method. Since these patterns are constructed in a manner 
similar to those of preceding problems, definite instructions are 
omitted. 

The principal feature of the problem consists in, first, finding 
the true line of intersection between the round pipe F and the 
square pyramid G. After this line has been found, the pattern 
for the round pipe is obtained in the usual way by the parallel¬ 
line method, as shown by half-pattern H. Draw the plan and 
elevation, and extend the sides of the pyramid in the elevation 
to the vertex at m. Next, in the plan, divide the circle into a 
convenient number of equal parts, as shown. These divisions 
should be so arranged that points will be located on the corners 
of the pyramid, as shown by 3 in the plan view. From the 
points 1 , 2 and 3 in the plan, draw vertical lines intersecting the 
side of the pyramid in the elevation at 12' and S'. From these 
points, draw horizontal lines to intersect vertical lines drawn 
from similarly numbered points in plan. A line traced thru 
these points will give the curved miter line in the elevation. 
This miter line is not necessary in developing the pattern, but is 
shown here to explain the method of projecting the miter line, 
no matter what form the pipe, or what pitch the pyramid may 
have. The pattern for one side of the base with the opening in 
the top is shown in D. 

To obtain the pattern, draw the center line m'c" equal in 


length to me' in the elevation. Thru C", at right angles to the 
center line, draw the line d’n' equal to dn in the plan view; then 
connect the points m'd' and n', which will give the full pattern 
for one side of the pyramid. A separate operation is necessary 
to find the outline of the opening in the upper portion of the 
pyramid, where it is intersected by the round pipe. 

From the points 1,2 and 3 of the plan, draw horizontal lines 
to the base line nd, as shown by 1-c, 2-b and 3-a. The points 
a, b and c in plan are now located on the line d'n' in the pattern 
by means of the dividers, as shown by a", b" and c". From 
these points erect vertical lines, making c"-l" in the pattern 
equal to c'-l' of the elevation, b"-2" in the pattern equal to C'-2' 
of the elevation, etc. A line traced thru these points will give 
the curved outline of the upper edge of the pattern. The pat¬ 
tern for the base is completed by taking a tracing of the straight 
side of the base in the elevation and placing it in position below 
the line d’n' in the pattern, as shown. Laps are allowed for 
seaming and flanging. 

Problem 113. A Round Pipe Intersecting an 
Elbow Miter 

In Figure 152 is shown the method for obtaining the patterns 
for a five-piece, 90 ° elbow intersected by a round pipe. It does 
not matter how many pieces the elbow may contain, or whether 
the pipe is placed in the center of the elbow or to one side, the 
principles explained in this problem apply in each case. The 
drawings in Figure 152 are shown one-third full size. Draw the 
full elevation of the elbow, as shown by the sections ABCD and 
E. Then, below the elevation, draw the profile of the elbow and 
the plan view of the round pipe; also its profile, shown by G'. 
Next, draw the elevation of the horizontal pipe in its desired po¬ 
sition, and draw the half-profile G in size equal to G' in plan. 









SPECIAL PROBLEMS 


165 


































































166 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


Divide both of the profiles G and G’ into the same number of 
equal spaces, as shown. From the various points on profile G’, 
draw horizontal lines intersecting the circle R in plan at A, 3-5 
2-6 and 1 -7. From these divisions on R, draw vertical lines in¬ 
tersecting the miter line mn in the elevation, from which points 
project lines across 
sections B and C, 
as shown. Now, 
from the various 
points on half-pro- 
file G, draw hori¬ 
zontal lines inter¬ 
secting similarly 
numbered lines in 
sections B and C, 
which have been 
projected from the 
plan. Aline 
traced thru these 
points will give 
the curved line of 
intersection be¬ 
tween the elbow and the round pipe, shown by 1', 2', 3', 
A', etc. 

The pattern for the horizontal pipe is developed by the 
parallel-line method in the usual manner, as shown at F in the 
drawing. A one-half pattern for section B is shown in B', and is 
developed in the following manner: Draw any vertical line, as 
a"b" in the pattern, upon which place the divisions gto7 equal 
to the divisions g to 7 in profile R in plan. Thru these points 
draw measuring lines, as shown. Then, measuring from the 
center line xx in section B, take the various distances to the 


points on the miter lines mn, a'b', and the curved miter line 1', 
2', 3 ', and place them on similarly numbered measuring lines in 
pattern B', measuring in each case on each side of the line a"b", 
as shown. A line traced thru these points will complete the re¬ 
quired pattern. The pattern for section C is developed in pre¬ 
cisely the same manner, as shown at C". 

Problem 114. Two-Branch Duct Fitting, Square 
to Rectangular 

We have chosen for treatment in this problem the construc¬ 
tion of a two-way elbow fitting from square to rectangular, 
which deals with a special case of square and rectangular duct 
construction occasionally encountered in the erection of blast 
systems for heating and ventilating. The large end is square in 
shape, and each branch is made in the form of a rectangular el¬ 
bow, the heel and throat being the quadrant of a circle. In 
Figure 153 is shown the elevation, plan and patterns of the fit¬ 
ting, which are drawn one-fourth full size, and we have for con¬ 
venience considered the square end of the larger pipe as the base 
or bottom of the fitting. 

Let FGCEab in the elevation represent the front and side 
views of the two branches of the two-way elbow. The profile of 
the rectangular end of one branch is shown by the shaded out¬ 
line FGHR. A side view of the other branch is shown by the 
circular outlines aC and bE, which are described from the cen¬ 
ters m and n. The width of the square end of the fitting is 
shown by the base line CE. 

After drawing the elevation, divide the quarter circles 1-11 
and 13-20 into equal spaces, as shown. The next step is to 
draw the plan view directly above the elevation, which shows 
the position of the two branches; also the miter or joint-line AB. 
The profile of the square end of the fitting is shown by ADBO in 



Problem 113. Round Pipe Intersecting 
an Elbow Miter. 







SPECIAL PROBLEMS 


167 













































































168 „ 


SHEET-METAL PATTERN DRAFTING AND SHOP PROBLEMS 


plan. The pattern for the outer side of each branch is shown in 
J. This pattern is simply a duplicate of the side view of one 
arm of the fitting, shown by abCE in the elevation; the arcs 
1-11 and 18-20 are struck 
from the centers m' and n\ as 
shown. Laps are added for 
seaming, as indicated by the 
dotted lines. The pattern for 
the inner side of both 
branches, which is constructed 
in one piece, is shown in P. 

The outline of one-half of this 
pattern, shown by a'-b'-8'-13', 
is a duplicate of a portion of 
the side view of one branch, 
shown in the elevation by a-b- 
8-13, to which laps have been 
added for seaming. 

The pattern of the heel of 
each branch is shown in K. 

It is developed by the paral¬ 
lel-line method as follows: From the points 1 to 8 in the eleva¬ 
tion, draw vertical lines, which intersect the miter line AB in 
plan. Now, draw the horizontal line 1-F in pattern K, and 
upon this line place the stretchout of the upper side of heel of 
the branch, the divisions 1 to 11 to F' being equal to similarly 
numbered divisions from 1 to a in the elevation. Thru these 


points, draw the usual measuring lines, which are intersected 
by horizontal lines drawn from corresponding points on the 
miter line AB in plan, extending point 8' to G', as shown. A 

line traced thru the intersec¬ 
tions from 1 to 8' will give the 
curved miter cut in the pat¬ 
tern for forming the two upper 
sides of the branches at an 
angle of 90 °. 

The pattern for the throat, 
shown in S, is rectangular in 
form. It can be laid out di¬ 
rectly upon the metal, the 
length being equal to the dis¬ 
tance from E to b in the ele¬ 
vation; the width is equal to 
the long side of the rectangu¬ 
lar ends of the fitting, shown 
by FGHR in the elevation. 
Edges are added to the pat¬ 
terns for seaming, as indi¬ 
cated by the dotted lines. Single edges have been allowed 
on patterns J and P, and double edges on patterns K and S, 
for joining the various pieces by means of a “Pittsburgh seam.” 
This is a quick method of seaming the corners of duct elbows 
without double seaming or riveting. It is easily constructed 
and makes a tight, rigid joint. 



Problem 114. Two-Branch Duct Fitting, Square to Rectangular. 







INDEX 


169 


INDEX 


A Page 

Angle, to bisect. 14 

to construct similar to given angle. 14 ' 

Angles, to transfer.14 

Ash or garbage chute.102 

Automobile measure.158 

B 

Ball, development by zones.150 

development by gores.152 

Base, for chimney top, rectangular to round.154 

Broken pediment.00 

Butt miter.• 44 

against a surface oblique in plan. 46 

C 

Center boot, round to oblong. 92 

Circle, definition of parts. 14 

to inscribe within a given triangle. 18 

to draw thru any three given points not in a straight line. 18 

Chimney top base, short rule for.154 

Chimney hood. 156 

Common bars in hipped skylight, to find true lengths of.147 

Conductor elbow, two-piece.26 

Chute, ash or garbage.102 

Conductor head. 54 

Conductor shoe, rectangular. 26 

Cone, envelope of and frustum. 66 

frustum of. 66 

intersected by a cylinder placed vertically. 78 

intersected by a vertical square pipe. 80 

intersected by a cylinder in a horizontal position 80 

intersected by a square pipe in a horizontal position 82 

the frustum of a cone intersecting a cylinder obliquely 82 

Cones, two intersections of, at an oblique angle.84 


Conic developments. 

Conical gutter outlet..... 

Cover, raised, oblong. 

rectangular, raised cover with rounded corners 


Page 
. 67 
68 
. 98 
100 


D 

Developments, parallel line. 

radial line. 

triangulation. 

triangulation, short method. 

Diagram of skylight bars. 

Drawing equipment. 

Drawing paper, preparation of. 

Drawing pencils. 

Duct fitting, square to rectangular. . 
Double-pitch skylight. 


E 

Eave trough miter. 

Egg-shaped oval, to describe. 

Elbow, conductor. 

Elbow fitting, two-way. 

Elbow miter, intersected by a round pipe 

Elbow profiles. 

Elbow, intersecting a round pipe. 

four-piece, 90°. 

five-piece, 60°. 

two-piece in an oblong pipe. 

three-piece, reducing, round to round. 

three-piece, oblong to round. 

development. 

three-piece transition, round to square 

two-piece in round pipe. 

three-piece, 90°. 


25 

64 

87 

103 

143 

9 

9 

9 

166 

138 


46 

18 

26 

128 

164 

40 

160 

38 

40 
28 

108 

110 

41 

no 

26 

40 

































































170 


E 

Page 

Ellipse, to describe. 18 

Elliptical flaring pan. 74 

Elliptical ship ventilator.134 

F 

Finial, for roof, square in plan. 52 

octagonal. 54 

Flaring pan, round.?. 72 

Flaring pan, elliptical. 74 

Flat skylight.136 

Flaring pan, rectangular, with folded corners.149 

Flaring pipe connection, vertical on one side. 92 

Florist’s sprinkler.156 

Frustum of a right cone. 66 

of a hexagonal pyramid. 68 

of a cone intersecting a cylinder obliquely. 82 

of a cone intersecting a cone obliquely. 84 

Funnel patterns. 74 

Fruit jar filler. 74 

Furnace boot, round to rectangular.112 

Furnace pipe fitting, round to oblong. 92 

G 

Gable molding. 50 

Gable molding with raked profile. 58 

Gable skylight.138 

Garbage chute.102 

Geometrical problems. 12 

Grocer’s scoop. 76 

Gutter, eave trough. 46 

molded face. 48 

molded roof. 48 

ogee. 48 

octagonal face. 48 

quarter circle. 48 

Gutter outlet, flaring. 68 

Gutter profiles. 49 


H 


Page 

Head, conductor. 54 

Hexagon, to construct. 16 

Hexagonal pyramid, the development of a. 68 

Hip bar, patterns for a hip bar in a skylight.145 

Hip bar, to find the true length of a.147 

Hipped skylight.140 

Hood for chimney top.156 

Hyperbola, to draw a. 20 

I 

Inside miters. 44 

Intersection of a cone and vertical cylinder. 78 

Irregular fitting, from round to oblong. 94 

Irregular flaring pipe connection, vertical on one side. 92 

Irregular flaring roof collar.103 

Irregular frustum of a cone. 66 

Irregular T-joint. 95 

Irregular T-joint, rectangular to round.126 

Irregular two-branch Y-joint.115 

Irregular two-branch fitting, round to square.118 

Irregular Y-fitting, round to square, mitering with vertical and hori¬ 
zontal connections.118 

J 

Jack bar, pattern for the top of jack bar in skylight.143 

Jack bars in hipped skylight, to find true lengths of.147 

Liquid measure, with funnel attached.158 

M 

Measure, automobile.158 

Measurements, skylight bars.147 

Miter, between an inclined molding and a horizontal return. 58 

between gable and horizontal moldings. 50 

between moldings of different profiles. 62 

eave trough. 46 

face miter at other than right angles. 50 
































































M 


Page 

oblique return. 46 

outside and inside. 44 

panel. 50 

square face miter, short method. 50 

square return, short method. 44 

Molding, to construct an ogee. 42 

Moldings, description of Roman moldings. 42 

Molded face gutter. 48 

Molded roof gutter. 48 

O 

Octagon, to construct. 16 

Oblique cone. 96 

Oblique return miter. 46 

Oblong pitched cover. 12 

Oblong raised cover with semi-circular ends. 98 

Octagon face miter. 50 

Octagonal face gutter. 48 

Octagonal pipe, intersecting a square pipe obliquely. 36 

Octagonal roof finial. 54 

Octagonal pyramid. 12 

Offset fitting, three-piece.106 

Ogee gutter. 48 

Outlet, tapering gutter. 68 

Oval, egg-shaped, to describe. 18 

P 

Panel miter. 50 

Parabola, to describe a. 20 

Pattern for cone and frustum. 66 

Parallel-line development. 25 

Pattern, for a funnel. 14 

for an irregular frustum of a cone. 66 

Patterns, for a double-pitch skylight.141 

for a flat skylight.139 

for a hipped skylight.143 

for a hip bar.145 


111 


P 

Page 

Pentagon, to construct a. 16 

Pencils, drawing.’•. 9 

Pediment, molding mitering on a wash. 56 

with normal profile used in an inclined molding. 60 

Pipe fitting, twisted. 94 

Pipe and roof flange. 28 

Pipe and roof flange to fit over the ridge of roof.. 30 

Pipe, octagonal, to fit over the ridge of roof. 30 

Pitched cover, oblong, with semi-circular ends._. 12 

Plates, drawing, preparation of. 12 

Practical geometry. 12 

Practical pattern-drafting. 24 

Pyramid, octagonal. 12 

rectangular. 10 

square. 66 

R 

Radial developments. 64 

Raked molding. 58 

Raised cover, rectangular, with rounded corners.100 

Raised oblong cover. 98 

Rectangular flaring pan with folded corners.-.149 

Reducing elbow, three-piece, round to round.108 

Register box, rectangular to round. 88 

Register box, rectangular to round, vertical on two sides. 90 

Ridge bar in hipped skylight, to find the length of.144 

Right cone, development. 66 

Return miter, short method. 44 

Roman moldings, to construct... 42 

Roman moldings, drawing plate. 43 

Roof collar, tapering.. 10 

square to round, base obliquely-inclined. 96 

irregular flaring.103 

round top and square base.106 

Roof flange. 28 

Roof finial, square. 52 







































































172 


R 

Page 

Roof finial, octagonal. 54 

Roof gutter.!.48 

Round pipe intersected by a pieced elbow.160 

Round pipe intersecting an elbow miter.164 

Round ship ventilator. .. .130 

Round ventilator head. 74 

Round ventilator with molded base.150 

Round ventilator with square base.162 

Rule for finding factors.148 

Rules for finding the true lengths of skylight bars.147 

S 

Scalene cone. 96 

Scoop, grocer’s. 76 

Scroll, to draw to a specified width. 22 

Short method, for developing face miters. 50 

for developing square return miters. 44 

triangulation.103 

Ship ventilator, round.130 

elliptical.134 

Skylight, flat.136 

double-pitch.138 

hipped.'.140 

Skylight bars, diagram of bars for a hipped light.143 

Skylight-bar intersections.146 

Skylight bars, rules for finding the true lengths of bars in a hipped sky¬ 
light.147 

Special problems.149 

Spiral, to construct a. 20 

Square face miter. 50 

Square pipe intersecting a cor.e. 80 

Square, to construct a. 16 

Square pipes, of same diameter, intersecting at an angle. 39 

Square return miter. 44 

Square roof finial. 52 

Square pyramid. 66 


Square ventilator 
Star, six-pointed. 


S 


Page 
. 52 
.100 


T 

Tapering forms. 64 

Tapering roof collar, for reef having a double pitch. 70 

Tapering square pipe intersecting a vertical square pipe. 76 

T-joint, pipes of same diameter. 32 

between pipes of different diameters. 33 

between pipes of same diameter at an angle. 33 

round pipes that intersect irregularly. 34 

square pipes intersecting at an angle. .*.. 34 

octagonal pipe intersecting a square pipe obliquely. 36 

rectangular pipe intersecting a round pipe obliquely. 38 

The helix. 22 

Three-branch fitting..120 

Three-way branch fitting.122 

Transition piece, square to square. 87 

rectangular to round. 88 

rectangular to round, vertical on two sides. 90 

round to oblong. 92 

rectangular to triangular. 94 

round to oblong, upper base at an angle. 94 

Transition elbow, oblong to round.110 

round to square.,.110 

Triangle, to inscribe within a circle. 18 

to draw, equal to any given triangle. 14 

equilateral, to draw, one side being given. 14 

Triangulation. 87 

Triangulation, short method.103 

Two-way elbow-fitting.128 

Two-branch duct fitting, square to rectangular.166 

Two-way transition branch, mitering with vertical and horizontal con¬ 
nections.124 

Twisted rectangular pipe. 94 



































































INDEX 


173 


Ventilator head, round. 74 

Ventilator, ship, with round mouth and base.130 

square in form.,. 52 

ship, with round base and elliptical mouth.134 

with molded base.150 

with square base.162 

W 

Wash, an inclined molding mitering on a. 56 

Watering pot, florist’s.156 


Y-joint, branches of same diameter. 33 

irregular, round to square.118 

irregular, two-branch.115 

two-branch, round to round. 116 

two-branch, round to square.116 

two-piece, tapering.114 

three-way.120 

three-branch. 122 

two-way transition branch, mitering with horizontal and vertical 
connections.:.124 


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BOOKS FOR TRADE INSTRUCTION 


ELEMENTARY FORGE PRACTICE 

By Robert H. Harcourt. A textbook and shop manual adapted for use 
by students in vocational, technical and high schools. It presents a series 
of over forty projects, covering the fundamental operations in forge prac¬ 
tice and their applications to commercial work. Full-page plates of work¬ 
ing drawings show the various steps in the development of each project. 
T he text accompanying each plate gives the needed explanations and help. 
In addition, there are chapters on tools, processes and materials, the matter 
being well selected and sufficient in quantity. In every way this is an ad¬ 
mirable text for student use. 

WOOD PATTERN-MAKING 

By Horace T. Purfield. A clear, concise treatise on the fundamental 
principles of pattern-making. It presents the best methods of construc¬ 
tion and those most easily understood by the student. It is not an anged 
about a course of problems, but may be used with any course. A practical 
text for high school, trade school, technical school, and engineering college 
students. Written by an experienced pattern-maker and teacher of pat¬ 
tern-making and kindred subjects. 

ELEMENTARY MACHINE SHOP PRACTICE 

By T. J. Talmateer. This book possesses the qualities needed in a text 
for use by students and apprentices. It offers problems, gives the needed 
explanations and just sufficient information. Unlike most books on this 
subject, it is neither a reference book nor a treatise; but it is a practical 
text, well adapted for school use. In many ways it is the best student text 
published on this subject. 

SCHOOL SHOP INSTALLATION AND MAINTENANCE 

By L. S. Greene. This book brings together selected facts known to en¬ 
gineers, and many more known to expert mechanics, and presents them 
from the standpoint of one who is acquainted with school shop conditions 
and the needs of teachers who are called upon to equip and take care of 
school shops. Power transmission, motors, installation of machinery, saw¬ 
fitting, brazing band-saws, belting, babbitting, etc., are included in the vol¬ 
ume. It will be a handy book for any shop teacher. 

CARPENTRY 

By Ira S. Griffith. A well-illustrated textbook for use in vocational 
schools, trade schools, technical schools, and by apprentices in the trade, 
presenting the principles of house construction in a clear and fundamental 
way. It treats of the “every-day” practical problems of the carpenter and 
house-builder from the “laying of foundations” to the completion of the 
“interior finish.” It meets every requirement as a textbook, and is also 


well adapted for reference use. It is well illustrated by photographs taken 
“on the job.” 

MECHANICAL DRAWING PROBLEMS 

By Edward Berg and Emil F. Kronquist. A direct and concise text 
adapted for students beginning mechanical drawing. It covers two years’ 
work, and contains 128 full-page plates—excellent examples of draftsman¬ 
ship. Text accompanies each plate, giving necessary facts and helpful 
hints wherever needed. The underlying principles of drafting are thoroly 
covered and the practical applications, which are abundant, have been 
most skilfully chosen and admirably presented. The plates tell what to do, 
almost at a glance, yet prevent mere copy work. Each problem tests the 
ability of the student to think and execute graphically, and unconsciously 
develops an excellent technique. 

PROGRESSIVE STEPS IN ARCHITECTURAL DRAWING 

By George W. Seaman. A textbook and practical hand-book, de¬ 
scribing and illustrating every successive step in drawing of floor plans, 
elevations, and various details for successful dwellings. Numerous plates 
illustrate details of doors, windows, mouldings, cornices, porches, etc. 
Architectural orders are shown in practical working forms; “single-line 
sketches” illustrate method of the practical designer in planning a house. 

ELECTRICAL CONSTRUCTION 

By Walter B. Weber. A combined textbook and shop manual presenting 
an elementary course for vocational schools. It contains 93 problems with 
accompanying diagrams and information. The problems are grouped ac¬ 
cording to difficulty and vary from simple magnetism to motors and gener¬ 
ators. A wonderful aid to the teacher and a valuable help to the student 
in securing definitely and firmly the correct principles and practices in¬ 
volved. 

MECHANICAL DRAFTING 

By H. W. Miller. (New edition revised and improved by R. K. Stew¬ 
ard.) A textbook for advanced students which presents drafting-room 
practice in practical textbook form. It is so written that it may be used 
with any course of exercises or problems, and supplements the instruction 
of the teacher in such a way as to reduce lecture work to a minimum. It 
is a direct and simple treatment of mechanical drafting, giving due con¬ 
sideration to the needs of the student, the beginning draftsman and the 
requirements of the best teaching methods. It is complete, yet condensed, 
and is well adapted for hand-book use by the student and draftsman. It 
is well illustrated and is bound in flexible binding, pocket size. A thoroly 
practical, modern textbook. 


PUBLISHED BY 

THE MANUAL ARTS PRESS, PEORIA, ILLINOIS 




































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